Differential Equations

DIFFERENTIAL

EQUATIONS

This book is in the

ADDISON-WESLEY SERIES IN MATHEMATICS

LYNN H. LOOMIS

Consulting Editor

DIFFERENTIAL

EQUATIONS

by

H. S. BEAR, JR.

Department of J.lf athematicsUniversity of Washington

ADDISON-WESLEY PUBLISHING COMPANY, INC.READING, MASSACHUSETTS

PALO ALTO • LONDON

Copyright © 1962

ADDISON-WESLEY PUBLISHING COMPANY, INC.

Printed in the United States of America

ALL RIGHTS RESERVED. THIS BOOK, OR PARTS THERE

OF, MAY NOT BE REPRODUCED IN ANY FORM WITH

OUT WRITTEN PERMISSION OF THE PUBLISHER.

Library of Congress Catalog Card No. 62-9398

PREFACE

This book is intended to serve as a text for a standard one-semester ortwo-term course in differential equations following the calculus. Theauthor has given more than the usual emphasis to the mathematical explanations, in the conviction that there is little value in learning techniquesby rote.

There is more material presented here than can be covered in two terms,and the material ranges from routine calculations to moderately sophisticated theorems. This variety allows the instructor a fair degree oflatitude in both the content and the level of the course to be taught.

The author would like to acknowledge his indebtedness to the University of Washington for the privilege of teaching from this text in apreliminary form, and to the many colleagues and students who havemade helpful suggestions.

January, 1962

v

H. S. B.

CONTENTS

CHAPTER 1. FIRST ORDER EQUATIONS

1-1 Introduction1-2 Variables separate.1-3 Geometric interpretation of first order equations.1-4 Existence and uniqueness theorem1-5 Families of curves and envelopes1-6 Clairaut equations

CHAPTER 2. SPECIAL METHODS FOR FIRST ORDER EQUATIONS

2-1 Homogeneous equations-substitutions2-2 Exact equations2-3 Line integrals .2-4 First order linear equations-integrating factors2-5 Orthogonal families2-6 Review of power series2-7 Series solutions

CHAPTER 3. LINEAR EQUATIONS

3-1 Introduction3-.:.2 Two theorems on linear algebraic equations3-3 General theory of linear equations .3-4 Second order equations with constant coefficients3-5 Applications .

CHAPTER 4. SPECIAL METHODS FOR LINEAR EQUATIONS

4-1 Complex-valued solutions4-2 Linear differential operators4-3 Homogeneous equations with constant coefficients4-4 Method of undetermined coefficients4-5 Inverse operators .4-6 Variation of parameters method

CHAPTER 5. THE LAPLACE TRANSFORM .

5-1 Review of improper integrals5-2 The Laplace transform5-3 Properties of the transform .5-4 Solution of equations by transforms

VII

1

138

121520

25

25313540464853

58

5859616874

83

838895

100105109

116

116119124131

viii CONTENTS

CHAPTER 6. PICARD'S EXISTENCE THEOREM

6-1 Review.6-2 Outline of the Picard method6-3 Proof of existence and uniqueness6-4 Approximations to solutions

CHAPTER 7. SYSTEMS OF EQUATIONS

7-1 Geometric interpretation of a system7-2 Other interpretations of a system .7-3 A system equivalent to M(x, y) dx + N(x, y) dy7-4 Existence and uniqueness theorems7-5 Existence theorem for nth order equations7-6 Polygonal approximations for systems .7-7 Linear systems7-8 Operator methods7-9 Laplace transform methods

INDEX .

o

140

140144147156

161

Hil165168171179184188194199

205

CHAPTER 1

FIRST ORDER EQUATIONS

1-1 Introduction. A differential equation is an equation containing an"unknown" function and its derivatives. If the unknown is a function ofone variable, the equation is called an ordinary differential equation, andif the unknown is a function of several variables so that the derivativesare partial derivatives, then the equation is a partial differential equation.We will be concerned exclusively with ordinary differential equations inthis book. We will study equations such as

4x3f"(x) - 2X[f'(X)]2 + f(x) - X2 = 0, (1)

. and ask what functions f satisfy the equation identically.As an example of how a differential equation arises naturally in a

physical problem, consider an object of mass m falling through the air.Suppose that the object is only slightly more dense than air (e.g., a balloon)and that the air flows smoothly around the object as it falls. In this casethe force due to air resistance is proportional to the speed. If we let s(t)be the distance the object falls in time t, then the speed is s'(t) and theacceleration is s"(t). The forces acting on the object are the force of gravity,mg, and the force of air resistance, ks'(t). According to Newton's law,force equals mass times acceleration, .and the motion is described by thedifferential equation

mg - ks'(t) = ms"(t). (2)

The highest order of the derivatives occurring in a differential equationis called the order of the equation. Thus a first order equation is anyequation of the form

G(x,f(x),f'(x)) = 0.

We customarily use the symbol y for the unknown function, and forsimplicity in writing we omit the independent variable from the expressions y(x), y'(x), etc. Hence a first order equation will appear as

G(x, y, y') = 0, (3)

and the second order equations (1) and (2) would ordinarily be written

4x3y" - 2x(y') 2 + y - x2 = 0,

mg - ks' = ms".1

(4)

(5)

2 FIRST ORDER EQUATIONS [CHAP. 1

A function is a solution of a differential equation on an interval I if thefunction satisfies the equation identically on I. Thus a solution of (3) isany function y such that

G(x, y(x), y'(x)) =: 0,

where U=:" is used to indicate that the equality holds for all values of xin some interval. The function f defined by f(x) = x 2 is a solution of (1)on the whole line, since

Similarly, the function set) = (mg/k)t is a solution of (2), since

mg - k (~g) =: m . O.

Notice that for any number c, the function set) = (mg/k)t + c is alsoa solution of (2). This illustrates the fact that generally speaking adifferential equation does not have just one solution, but an infinite familyof solutions. This fact is familiar from the study of indefinite integrationi.e., the study of differential equations of the form

y' = f(x).

We will say that two equations are equivdJent if any function whichsatisfies either equation also satisfies the other. That is, two equationsare equivalent if they have the same sets of solutions. The following equations are easily seen to be equivalent:

The equations

y' + y = X,

xy' = x2 - xy,

eY(y' + y - x) = O.

yy' + y2 = xy,

(1 + y)(y' + y - x) = 0

(6)

(7)

(8)

are not equivalent to the equations (6), since each has a solution whichis not a solution of (6) (Problem 4).

PROBLEMS

1. Show that for any numbers CI and C2 the function set) = (mg/k)t + C1 +C2e-(k/mlt is a solution of (5).

2. For what values of c is x 2 + c a solution of (4)?

1-2] VARIABLES SEPARATE 3

2. c = 0

3. (a) y = crx + C2(c) Y = sin-rx+c

3. Solve the following equations.

(a) y" = 0 (b) y' 2 (c), 1

= sec x y =VI - x 2

(d) y"x+I (e) y'

1 (f) , 1=--

= 4+ x 2y =---

X x 2 - 9

4. Show that neither (7) nor (8) is equivalent to the equations (6). Are (7)and (8) equivalent?

5. Show that the functions given below are solutions of the correspondingdifferential equations. Find one more solution of each equation.

(a) x 2y' = xy - x, y = 2x + 1 (b) y" + y' = 0, y = 3 - 2e-X

(c) (x + I)y' = -y, Y = Ij(x+ 1) (d) y = (x + I)y', Y = x + 1(e) y" + y = 0, y = cos x

ANSWERS

(b) y = tan x + c(d) y = tx2 + x In Ixl + crx + C2

-1 X Ix - 31(e) y = t tan 2+ c (f) y = i In x + 3 + c

4. (7) has the solution y = 0, (8) has the solution y = -1; no.

5. (a) y = cx + 1 (b) y = C1 + C2e- X

(c) y = cj(x + 1) (d) y = c(x + 1)(e) y = Cr cos x + C2 sin x

1-2 Variables separate. For any equation of the simple form

y' = f(x),

the solutions are obtained by integrating both sides. Thus

y = ff(x) dx + c.

(1)

(2)

That is, to solve (1) we are required to find each function y whose derivative equals f. If F is any function such that F'(x) = f(x), then anyfunction y satisfying (1) differs from F by a constant. Therefore, allsolutions of (1) must have the form

y(x) = F(x) + c. (3)

Conversely, any function y of the form (3) obviously satisfies (1), andhence (3) characterizes the family of solutions of (1).


Aside from the familiar equations of the form (1), the simplest differential equations are those of the form

g(y)y' = f(x). (4)

We will say that the variables separate in a differential equation if it isequivalent to an equation of the form (4). We agree that theequation

g(y) dy = f(x) dx (5)

is equivalent to (4); i.e., we define y to be a solution of (5) if and only if yis a solution of (4).

As an example of an equation in which the variables separate, consider

(6)

This is equivalent to

(4y 3 + l)y' = 1 - 3x2

and hence by definition to

(4y 3 + 1) dy = (1 - 3x2) dx.

If we integrate both sides of (7), we obtain

y4 + y = x - x 3 + c.

(7)

(8)

Two questions arise concerning this process. First, in what sense is (8) asolution of (6), since (8) is not a family of functions but a family ofequations? The answer is that the solutions are the fu~ctions y whichsatisfy an equation of the form (8); i.e., the functions y which are definedimplicitly by one of the equations (8). In many cases it is not possible towrite explicit formulas for the solutions, and we regard the differentialequation as solved if we can characterize the solutions in a manner notdependent on their derivatives. The second question is whether the familyof equations (8) actually does characterize the solutions of (6); does everysolution of (6) satisfy one of the equations (8), and is every functionsatisfying one of the equations (8) a solution of (6)? It is by no meansobvious that the formal process of integrating both sides of (7) solves thedifferential equation; this process is justified by the following theorem.

THEOREM 1. If f and g are continuous, and F' = f, G' = g, and y isany solution of g(y) dy = f(x) dx, then there is a constant c such thatG(y(x») = F(x) + c. Conversely, any differentiable function y whichsatisfies G(y) = F(x) + c, for some c, is a solution of the equationg(y) dy = f(x) dx.


Proof. The second assertion, that any function satisfying an equationG(y) = F(x) + c is a solution of the equation g(y)y' = f(x), is immediatefrom the chain rule for differentiation. Hence, we need only prove thatthere are no other solutions.

Let y be any solution of

so that

g(y) dy = f(x) dx,

g(y(x)y'(x) = f(x)

(9)

(10)

(12)

for all x. Let H(x) = G(y(x), where G is any antiderivative of g. Thenby the chain rule and the identity (10) we get

H'(x) = G'(y(x)y'(x) = g(y(x)y'(x) = f(x) = F'(x). (11)

That is, Hand F have the same derivative. It follows that there is aconstant c such that H(x) = F(x) + c, and hence that G(y(x) = F(x) + c.

EXAMPLE 1. (A) 2yy' = eX. G(Y

By the theorem, the solutions are just those functions y which satisfyy2 = eX + c, for some number c.

Except for the possibility y = 0, this equation has the same solutions asy-2y' = x2. We check that y = 0 is a solution, and integrate to solvey-2y' = x2. The solutions of this equation are given by the equations_y-l = !x3 + c, or

-3y = x 3 + 3c'

Since c is arbitrary, this family of functions can be written

-3y = x3 + c·

The solutions of i/ = X2y2 are therefore the function y = 0 and the

functions (12).

(C) x dx + y dy = O.

Integration gives !x2 + ty2 = c, which is the same family of equationsas x 2 + y2 = C. Note that no real-valued function y satisfies this lastequation unless c > 0, so the solutions of (C) are also characterized bythe equations x2 + y2 = c2

.


(D) x + 1 dx = (x z + 1) In Iyl dy.y

Multiplying both sides by y, and dividing by x Z + 1, we obtain

x + 1X Z + 1 dx = yIn Iyl dy.

Integrating the left side above, we get

f Cz ~ 1 + xZ ~ 1) dx = ! In (xz+ 1) + tan-

1 x.The right side can.be integrated by parts to give

f yIn Iyl dy = !yz In Iyl - f !yZ ~ dy

= !yZln Iyl - h Z.

The solutions of (D) are therefore the functions y defined implicitly bythe equations

2 In (X Z+ 1) + 4 tan -1 x = 2yz In Iyl - yZ + c.

A type of equation more general than that in which the variablesseparate is the exact equation

1)F(x, y) = 0, (13)

orFx(x, y) + Fy(x, y)y' = O.

We agree that the following equation is equivalent to (13) or (14).

Fx(x, y) dx + Fy(x, y) dy = O.

As an example of an exact equation we consider

2x - 2xyy' - yZ = O.

This equation can be written

(14)

(15)

(16)

(17)

A function y is a solution of (16)-that is, by (17), a function such that

D(xZ - x[y(x)j2) == 0,


, 8. y dx + x dy = 010. (y2 + 2x) dx + 2xy dy = 0

if and only if there is a constant c such that

In conventional form, therefore, the solutions of (15) are given by theequations

An analysis similar to the foregoing, and to the proof of Theorem 1,shows that the solutions of (13) or (14) or (15) are those functions iJ whichsatisfy an equation of the form F(x, y) = c (Problem 12).

EXAMPLE 2. (A) xy' + 2yy' = x - y.

This equation can be written

(xy' + y) + 2yy' - x = 0,

or

The solutions are therefore given by

xy + y2 - !x2 = c.

(B) (y2 + 3x2) dx + 2xy dy = 0.

This is equivalent to

(y2 + 2xyy') + 3x2 = 0,

or

The solutions are determined by the equations

xy2 +'x3 = c.

Exact equations are studied in more detail in Section 2-2 and Section 2-3.

PROBLEMS

Solve the following equations.

-'1. y' = xjy2. xy dy + (x 2+ 1) dx = 03'. (1 - y2) dx - xy dy = 0 4. xy + VI + x2 y' = 0

5. (y + 1) dx + (y - 1)(1 + x2) dy = 06. (2x + l)y' + y = 0

7. y' - x 2 = x 2y9. x2 dy + 2xy dx = x2 dx


11. Show that any equation of the form f(x) dx + g(y) dy = 0 is exact.12. Show that the solutions of (14) are the functions y defined implicitly by

an equation of the form F(x, y) = c.13. Show that the equation y' + y = 0 becomes exact if multiplied by eX,

and solve it.14. Show that the equation y - xy' = 2y3y' becomes exact if multiplied by

1/y2, and solve it.15. Find each positive function whose derivative is the reciprocal of the

function.16. Find all functions such that the square of the function plus the square of

the derivative is a given constant A 2.

17. Find all functions such that the derivative is the square of the function.18. Find all functions with derivative one more than the square of the

function.

ANSWERS

1. y2 - x2 = c 2. x2 + y2 + 2ln Ixl = c3. x2(1 - y2) = c 4. VI + x2 + In Iyl = c, y 0

5. tan-1x+ y - 2lnjy+ 1!- = c, y = -16. (2x + 1)y2 = c7. 3ln 11 + yl = x3 + c, y = -1

8. xy = c 9. 3xy2 = x3 + c10. xy2 + x2 = c 13. y = ce-:r:14. x/y = y2 + c, y = 0 15. y V2x + c16. y Asin(x+c) 17. y = -i/(x+c), y 018. y = tan (x + c)

1"';3 Geometric interpretation of first order equations. In Section 1-1we saw how a differential equation arises in the attempt to solve thephysical problem of a falling body. Consider now a geometrical problem.What functions, with graphs in the first quadrant, have the property thatfor every tangent line the point of tangency bisects the segment cut off bythe coordinate axes (Fig. I-I)? For such a curve, the tangent at the'point (x, y) must have slope -2y/2x = -y/x. Hence any such furiction'must satisfy the differential equation

y' = -y,x

y> 0, x > 0, (1)

and conversely, a function satisfying (1) satisfies the conditions of theproblem. The solutions of (1) (Problem 8, Section 1-2) are the functionsy = c/x, c > 0.

Any first order differential equation y' = F(x, y) admits a geometricinterpretation similar to that illustrated in the problem above. The equation y' = F(x, y) prescribes a tangent line at each point of the plane:

1-3] GEOMETRIC INTERPRETATION 9

(0,2y)

the line through (x, y) whose slope is F(x, y). The question posed by thedifferential equation y' = F(x, y) is: What functions y have the tangentsprescribed by the function F at each point of their graphs? We canvisualize the situation by drawing short segments of the tangent linesgiven by the equation at various points of the plane. Figure 1-2 showssuch a "tangent field" for the equation

y' ~ y - x, (2)


\.... I I

\. \. \. / I

\. \. \. / I

, '

\ \

, \ \

, \ \

\

/ / I

y=x2

I I

, '

\\/ / - , \

\

/ / , \ \/ \ \

/ / / , \ \ \

/ \\

/ I , \ ,/

/ ' , \\

// \

,Y= -X2 / I

/ \\ \

I / / \ \ ,/ / / \ \ ,/ / / \ \

,

I III -:. --;. ~ -;-;,,::./:"'... ~~ ... ~~ ~ ~"''''',,, \.

1/ /1//// // /,. ,,"'" "",'\. ,\'

1 1 //11

1/ 1// '" ,,' \. ,\ \\

II / , " "I A , , \. "

/1 / " !'o. \. \. "

I II /" ",' \.I I.. ... \. \.

FIG. 1-3. Tangent field for y'

with graphs of the solutions y = x + 1 + eX and y = x + 1 - eX. Aneasy way to construct a graph of the tangent field is to draw all the segments with a given slope at the same time. This amounts to graphing thecurves F(x, y) = c [the lines y - x = c for (2)] and marking segmentswith slope c at intervals along the curves. Notice that the linear solution,y = x + 1, can be found by inspection of the tangent field for (2).

Figure 1-2 suggests that there is a solution, and only one solution,through any given point in the plane. The solutions of (2) are thefunctions

y = x + 1 + cex, (3)

and it is easy to check that for any point (a, b), there is exactly one number c such that b = a + 1 + cea

. In other words, there is exactly onesolution curve through each point of the plane. It is true in general that

1-3] GEOMETRIC INTERPRETATION 11

if F is continuous, there is a solution to y' = F(x, y) through each point(a, b). The solution through a given point is not necessarily unique withoutfurther assumptions on F.

Let us examine in some detail the equation

(4)

The tangent field for (4) with the tangent segments drawn along thehyperbolas 2~ = c is shown in Fig. 1-3. On any interval on which asolution is never zero, we can separate variables in (4) and obtain theequivalent form

(5)

According to Theorem 1, Section 1-2, we obtain an equivalent equationby integrating both sides to get

or(X

4/

3 > c). (6)

Thus we have shown that any solution of (4) must have the form (6) onany interval on which it is nonzero.

The curves (6) which intersect the x-axis (c 2: 0) are tangent to thex-axis at the point of intersection, and hence satisfy (4) at this point(since y = y' = 0). The function identically zero isa solution of (4) onany interval. Hence any function whose graph follows one of the curves(6) to the x-axis, and continues along the x-axis until it leaves by anotherof the curves (6), isa solution of (4). There is a solution through everypoint, and there are infinitely many solutions through any point on thex-axis. As an example, the function

y(x)

x ~ 0

o < x < 1

1 ~ x

is a solution of the differential equation. The functions y(x) = x 2,

y(x) = -x2, y(x) = 0 are other solutions through the origin.

PROBLEMS

1. Find all curves such that for any point on the curve, the area bounded bythe ordinate line to the point, the x-axis, and the tangent line at the 'point is t.Find the one of these curves through (0, 1).

2. Construct a tangent field for y' = 1 + x + y. Find a linear solution byinspection of the tangent field and check it in the equation.


3. Consider the curves such that the tangent at each point (x, y) of the curveintersects the y-axis at (0, -y). Draw the tangent field in the upper half-planefor the implied differential equation. [Hint: The tangent segment at eachpoint on y = 1 must point toward (0, -1), etc.] Write the differential equation,find all such curves, and graph several of them on the tangent field.

4. (a) Construct a tangent field for y' = 2VY. (Note that necessarilyy ~ °and y' ~ 0.)

(b) Find all solutions of y' = 2vy as in the treatment of Eq. (4) in thetext.

(c) How many solutions are there through each point on the x-axis? Eachpoint off the x-axis? Explain.

5. Describe the solutions of y' = 3xy l/3.

ANSWERS

1. y = 1j(x + c), y 1j(x.+ 1)2. y = -x - 23. y = cx2

4. The solutions are y = 0, and the functions Yc, where yc(x) = °if x :s; cand yc(x) = (x - c)2 if x ~ c. There are infinitely many solutions througheach point on the x-axis, and one solution through each point in the upper halfplane.

5. The solutions are those continuous functions whose graphs lie always onone of the curves y = (x2 + c)3/2, Y = 0, or y = -(x2 + c)3/2.

1-4 Existence and uniqueness theorem. The examples in the precedingsection lead us to ask for a general statement about the existence anduniqueness of solutions of the first order equation

y' = F(x, y). (1)

The question we ask is this: If the function F satisfies certain conditionsnear a point (a, b), is there a solution curve passing through (a, b)? If SO,

is there exactly one solution through (a, b)? In other words, we wouldlike to know that there is exactly one function y such that yea) = b, andy'(x) == F(x, y(x)) on some interval around a.

There are several reasons for asking questions such as these. First ofcourse is the fact that before we look for solutions of a differential equation, we would like to be assured that "there are solutions to be found.The uniqueness part of the theorem is also of more than academic interest.We will illustrate below how the uniqueness of the solution through eachpoin"t~can be used to show that a given family of solutions contains allsolutions.

We state here one convenient form of an existence and uniquenesstheorem for Eq. (1). The hypothesis we use is the continuity of F andFy = aFjay near a given point (a, b). In geometric terms, the fact that

1-4] EXISTENCE AND UNIQUENESS THEOREM 13

F is continuous means that the tangent field determined by F changesdirection smoothly as the point (x, y) moves in the plane near (a, b). Wewill see in Chapter 6 that the continuity of F y rules out the sort of behaviorillustrated in Fig. 1-3, where solution curves are tangent to each other,and there can be several solutions through a given point.

THEOREM 1. If F and Fyare continuous on some square S centered at(a, b), then there is a function y defined on some interval I around a suchthat y(a) = b and y'(x) = F(x, y(x) for all x in I. Moreover, if 9 isany function such that g(a) = band g'(x) = F(x, g(x) for x in I, theng(~) = y(x) for x in I.

This theorem is proved in Chapter 6.Note that the existence theorem is local in character. If F is well behaved

near (a, b), then there is, on some interval I around a, a solution whosegraph goes through (a, b). The interval I may be small, even if F and Fyare continuous on the whole plane (see Problem 1).

EXAMPLE 1. Consider the equation (cf. Fig. 1-2)

y' = y - x. (2)

This is of the form (1), with F(x, y) = y - x, and Fy(x, y) = 1. Thefunction's F and Fyare continuous everywhere, so there is a unique solution through each point (a, b). It is easy to verify (Problem 2) that forany number c, the function

(3)

is a solution of (2). For any given point (a, b) there is a number c suchthat

b = cea + a + 1.

That is, there is a solution from the family (3) through any given point.Since by Theorem 1 there is only one solution through any point, thesolution through a given point must be the solution from the family (3).In other words, the family (3) contains all solutions of (2).

EXAMPLE 2. Consider the equation (cf. Fig. 1-3)

(4)

Here F(x, y) = 2x1/ 3yl/3, and Fy(x, y) = ~Xl/3y-2/3. The function F iscontinuous everywhere, but Fy is continuous only at points (x, y) withy ~ O. There is a solution of (4) through every point (a, b), but notnecessarily a unique solution. If we restrict our attention to a square


wholly above or below the x-axis, then there is in the square a uniquesolution curve through each point.

There are existence theorems analogous to Theorem 1 for equations oforder higher than one. For example, for the second order equation

y" = F(x, y, y') (5)

the following is true. If F is a function of three variables which is continuous near the point (a, b, c) and has continuous partial derivatives withrespect to the second and third variables, then there is a unique function y

defined on some interval around a which satisfies the differential equation

y"(X) == F(x, y(x), y'(x)),

and satisfies the initial conditions

yea) = b, y'(a) = c.

PROBLEMS

1. Solve the equation y' = 100 + y2. Show that even though F(x, y) =

100 + y2 and Fy(x, y) = 2y are continuous everywhere, no function is a solution of the equation on any interval longer than 11'/10.

2. Verify that each function in the family (3) is a solution of the Eq. (2).3. (a) Use Theorem 1 to show that y' = 1 + x - y has exactly one solution

through each point.(b) Verify that y = x + ce-X is a solution for every number c.(c) Show that for every point (a, b) there is a number c so that the solution

y = x + ce-X satisfies y(a) = b.(d) Conclude from (a), (b), and (c) that the family y = x + ce-X is the

family of all solutions of y' = 1 + x - y.4. Proceed as in Problem 3 to show that y' = 3x3 - 2xy has a unique

solution through each point, and that y = ce-x2 + x2 - 1 is the family of allsolutions.

5. Solve the following equations. Write the equations in the form y' = F(x, y)and show by Theorem 1 that there is a unique solution through the given point(a, b). Find the particular solution through the given point.

(a) x dy + y dx = x 2 dx; (a, b) = (1, 0)(b) (1 + 2x2) dy - x(1 + y2) dx = 0; (a, b) = (0, 1)(c) (l/x) dy - 4Vl + y dx = 0; (a, b) = (2,3)(d) dy = 2xy dx; (a, b) = (1,0)

6. Solve the following second order equations and find the solution whichsatisfies the given initial conditions.

(a) y" = y'; y(O) = 2, y' (0) = 1(b) 2y'y" = 1; y(l) = 5, y' (1) = 2(c) xy" + y' = x2; y(l) = 1, y'(I) = 1 [cf. Problem 5(a)]

1-5] FAMILIES OF CURVES AND ENVELOPES

ANSWERS

15

1. y = 10 tan (lOx + c)

5. (a) 3xy = x3 + c; c = -1(b) 4tan-1 y = In (1 + 2x2 ) + c; c = 71"

(c) vI + y = x2 + c and y = -1; c = -2(d) y = cex2

; c = 0

6. (a) y = CleX + C2, Y = eX + 1(b) y = t(x + Cl)3/2 + C2, Y = t(x + 3)3/2 - !(c) y = tx3 + ci In Ixl + C2, y = t x3 + tIn Ixl + ~

1-5 Families of curves and envelopes. We have seen that the set ofsolutions of a differential equation can frequently be expressed as a singleformula with one or more arbitrary constants, or parameters. We willconsider now the problem converse to that of solving a differential equation; namely, given a family of functions, is there a differential equationfor which this family is the set of solutions? In other words, can we characterize a family of functions with a single differential equation?

If we have a family of functions

y = F(x, e),

then each function in the family (1) must also identically satisfy

y' = Fx(x, e).

(1)

(2)

If we can eliminate e from these equations-that is, derive from (1) and(2) an equation

G(x, y, y') = 0,

then this is a differential equation which is satisfied by every function inthe family (1).

Consider the family of functions

y = 1 + ex.

Each of the functions (3) satisfies y' = e, and hence the equation

y = 1+ xy'.

(3)

(4)

Equation (4) is therefore a differential equation whose solutions containall the functions (3). By separating variables and solving, one shows that(4) has no solutions other than those from the family (3).

A two-parameter family of functions will be represented by a secondorder equation. Differentiating the formula

(5)

16 FIRST ORDER EQUATIONS

FIGURE 1-4

[CHAP. 1

twice, we see that every function (5) satisfies the three equations

y = CIX + C2ex,

y' = Cl + C2eX,

y" = C2ex.

(6)

We can solve the third equation for C2, then the second for Cl, and substitute these values in the first equation to obtain

y = (y' - y")x + y". (7)

We will be able to show in Chapter 3 that there are no other solutionsof (7) and hence that (7) does characterize the family (5).

It is not always the case, as we show next, that the differential equationwe derive from a family of functions has only solutions from this family.Consider the family of lines tangent to the curve y = x 2 (Fig. 1-4). Thetangent line at (e, e2

) has the equation y - e2 = 2e(x - c), or

(8)

To obtain the differential equation of the family (8), we differentiate andget y' = 2e. The differential equation of (8) is, therefore,

(9)

The equation (9) prescribes a slope for a solution curve at each point ofthe plane. At each' point (x, x 2

), the curve y = x 2 has the same slope asits tangent line, which is a solution of (9) by virtue of the way (9) was

1-5] FAMILIES OF CURVES AND ENVELOPES

FIGURE 1-5

17

derived. Hence y = x 2 is a solution of (9), and (9) does not characterizethe family of lines (8).

The situation encountered above occurs whenever a family of curveshas an envelope. We will say that the curve y = q,(x) is an envelope ofthe family of curves y = F(x, c) if at each point on y = q,(x) there is acurve· from the family which is tangent there (Fig. 1-5). That is, for eachpoint (xo, q,(xo») there is a value of c, c = c(xo), such that y = q,(x) andy = F(x, c(xo») are tangent at (xo, q,(xo»). We also suppose that c'(x)exists and c'(x) ~ O. With these assumptions, we can give a simple testfor determining whether a family y = F(x, c) has an envelope, and wecan write a formula for the envelope if there is one. The facts are statedin the following theorem.

THEOREM 1. The family of curves y = F(x, c) has the envelope y = q,(x)if and only if there is a function c such that Fc(x, c(x») == 0, andq,(x) = F(x, c(x»).

To restate the theorem: The envelope, if any, is the curve y = F(x, c(x»),where c is defined implicitly by the equation Fc(x, c) = O. The equationof the envelope is obtained by eliminating c between the equations

y = F(x, c), o = Fc(x, c).

Proof. Assume that y = q,(x) is an envelope of the family y = F(x, c).Let c be the function such that y = q,(x) and y = F(x, c(xo») are tangentat (xo, q,(xo»), for each xo. Since the curves y = q,(x) and y = F(x, c(xo»)meet at (xo, q,(xo») , we have q,(xo) = F(xo, c(xo»), for each Xo; that is, we

-- have the identityq,(x) = F(x,c(x»).

The condition that the curves be tangent implies that for each Xo

(10)


y=(X-C)3

FIGURE 1-6

[CHAP. 1

Hence we have the identity

ep(x) = Fx(x, c(x»).

If we compute ep' from (10), we get

ep'(x) = Fx(x, c(x») + Fc(x, c(x»)c'(x).

Comparing (11) and (12) we have, since c'(x) rf 0,

Fc(x, c(x») = 0.

(11)

(12)

(13)

Formulas (13) and (10) are the assertions of the theorem, if we assumethat there is an envelope.

The arguments above can be reversed to show that (13) and (10) aresufficient for y = ep(x) to be an envelope of the family y = F(x, c). Thispart of the proof is left as a problem (Problem 8).

EXAMPLE 1. Consider again the family

Equation (13) for this example is

2x - 2c = 0,

which defines the function c(x) = x.[Formula (10)]

The envelope is therefore

EXAMPLE 2. Consider the family (Fig. 1-6)

1-5) FAMILIES OF CURVES AND ENVELOPES 19

Partial differentiation with respect to c gives [Eq. (13)]

0= 3(x - C)2,

and hence c = x. There is, therefore, the envelope

y = (x - x) 3 = O.

For a family of curves given in the form

G(x, y, c) = 0 (14)

(15)Gc(x, Y, c) = o.

it can be shown that the envelopes, if any, are obtained by eliminating cfrom (14) and

EXAMPLE 3. (x - C)2 + y 2 = 1.

We eliminate c between the equations

2(x - c)(-I) = 0

to obtain the equation of the envelope curves, y2 = 1.

Summary. To firid the differential equation of an n-parameter familyof curves, differentiate the given equation n times and eliminate all theconstants.

To find the envelope of a one-parameter family of curves, eliminate theparameter between the given equation and that obtained from it by partialdifferentiation with respect to the parameter.

PROBLEMS

1. Solve Eq. (4) and show that there are no solutions other than thefunctions (3).

2. Find the differential equations of these families.·

(a) y = c + cx (b) y = C1 + c2e~

(c) y = x 2 + cx + c2 (d) y = C1 + C2X + C3X2

3. Considering (6) as three simultaneous linear (algebraic) equations in C1

and C2, the condition that the equations be consistent is that the following determinant be zero:

~x e y

1 e~ y' = O.

o e~ y"

Expand the determinant to obtain (7).


(b) y" = y'(d) yll/ = 0

4. Proceed as in Problem 3 to find the differential equation of the familyy = ele-x + e2ex - x.

5. Show that each of the parabolas y = x2 + ex + *(1 - e)2 is tangent toy = x. Find a first order differential equation for the family and verify thaty = x is also a solution.

6. Find the envelope of the family y = eX-

C + e. Graph several of the curvesand the envelope. Find the differential equation of the family and verify thatthe envelope is also a solution.

7. Find the envelope of the family y = 3e2x - 2e3 .

8. Show that if e is a differentiable function such that e' (x) ;;c 0 andFc(x, e(x») = 0, then y = F(x, e(x») is an envelope of the family y = F(x, e).(See the proof of Theorem 1.)

9. Show that there is no envelope to the family of solutions of g(y)y' = f(x).[Hint: Write the family of solutions and show that (15) is impossible.)

10. Solve 4y = y'2 by separating variables. Show that the family of solutionshas an envelope which is a solution. Reconcile these facts with Problem 9.

11. Show that every solution of (9) everywhere agrees with one of the functions (8), or with y = x2 . [Hint: Differentiate (9) and examine the resultingsecond order equation. Show that either y' is constant, and y is in the family (8),or y = x2 + e with e necessarily zero.)

ANSWERS

2. (a) y = y' + xy'(c) y = 3x2 - 3xy' + y'2

4. y" - Y - x = 05. y = x2 + (y' - 2x)x + HI - y' + 2X)26. y =. x + 1; Y = y' + x - In y'7. y = x3

10. y = (x - c)2, Y = 0; the separated form is not equivalent to theoriginal.

1-6 Clairaut equations. Now we consider the differential equation of aparticularly simple family of curves-a family of straight lines. Thedifferential equation of such a family turns out to have an easily recognizable form. We have, therefore, a class of differential equations forwhich a family of solutions can be found by inspection.

The families of lines we consider are those which can be written in theform

y = ex + gee). (1)

Differentiating (1), we obtain y' = e, and hence each of the functions (1)must satisfy

y = xy' + g(y'). (2)

1-6] CLAIRAUT EQUATIONS 21

Any equation of the form (2) is called a Clairaut equation. For any suchequation, we can write down by inspection the one-parameter family (1)of solutions. There may be solutions of (2) not contained in the family (1),and we know that any envelope of (1) would be such a solution.

EXAMPLE 1. The equation

y = xy' - i(y')2

is a Clairaut equation and therefore has the solutions

This family of lines has the envelope (cf. Section 1-5) Y = x 2, which is

also a solution.

EXAMPLE 2. y = xy' + 2 + Y'

This equation is of the form (2), so we can write by inspection the solutions

Y = ex + 2 + e.

Differentiating this equation with respect to e we get 0 = x + 1, whichis obviously not satisfied by any function e. The family of solutions hasno envelope.


(y - y'x) 2 = 2y' + 1

is essentially a Clairaut equation and has the solutions

(y - ex) 2 = 2e + 1.

To check for an envelope, we differentiate with respect to e and obtain

2(y - ex) (-x) = 2, 1 + xyc= .x 2

The family has the following envelope, which is a solution of the differential equation,

We have left unresolved the question of whether there are solutions of(2) other than the family (1) and its envelopes. We can obtain othersolutions of (2) by piecing together functions from the family and the


envelope at points of tangency. For example, the function

{

0,

y = 2x x2

, 2,

x ~ 0o ~ x ~ 1

1 ~ :t

(3)

is a solution of the equation of Example 1. The question is, therefore,whether every solution of (2) must everywhere coincide either with oneof the functions (1), or with an envelope of the family (1). We show inthe following theorem that this is the case.

THEOREM 1. If Y is any solution of y = xy' + g(y'),then y ~·s everywhereequal either to one of the functions y = ex + gee), or to an envelope ofthis family.

Proof. If y satisfies the equation (2), then by differentiating andsimplifying, we see that y also satisfies the identities

y' = xy" + y' + g'(y')y",

y"[x + g'(y')] = o.

On any interval on which y" = 0, y' is some constant c, and from (2) weconclude that

y = ex + gee).

If the other factor in (3) is zero,

x + g'(y') = 0,

then there is a function e = c(x) satisfying

(4)

(5)

(6)ao = ae [ex + gee)] = x + g'(e);

in fact, the function e(x) = y'(x) satisfies (6), by (5). Equation (6) isjust the condition of Theorem 1, Section 1-5, that the family (4) have anenvelope. By the theorem, the envelope function ep is given by

ep(x) = e(x)x + g(e(x)) = xy'(x) + g(y'(x)). (7)

Comparing (7) and the differential equation,

y = xy' + g(y'),

we see that the solution y is an envelope of the family (4), if y' satisfies (5).To summarize, a function y which satisfies (2) is one of the functions

y = ex + gee) on any interval where y" = 0, and is an envelope of thisfamily where x + g' (y') = O.

1-6] CLAIRAUT EQUATIONS 23

The technique of the proof above extends to equations other than theClairaut equation. Suppose we have any first order equation which canbe solved for y-that is, written in the form

y = F(x, y'). (8)

The solutions of (8) are given explicitly by the equation itself, once thederivatives y' are known. If we differentiate (8), we obtain a second orderequation involving only x, y', and y". Such an equation can be regardedas a first order equation in y'. If this derived equation can be solved fory', then the solutions of (8) are obtained by substituting these values of y'in (8).

EXAMPLE 4. Consider again (cf. Example 1) the equation

(9)

Differentiation gives the equation

y' = xy" + y' - !y'y",

y"[x - !y') = O.

Therefore we must have y' = c (if y" = 0), or y' = 2x. Substitutionin (9) gives all solutions,

or

EXAMPLE 5.2

y = xy' +x

(10)

The variables cannot be separated in (10), nor is it exact or a Clairautequation. For any solution y, however, we must have

21" = xy" + y' - - ,v x2

y" = 2x-3.

Therefore y is a solution if and only if

y' = -X-,2 + e,

and the solutions are, by substitution in (10),

y = x(-x-2 + e) + ~ = ex +!.x x


PROBLEMS

[CHAP. 1

1. Find all solutions, including envelopes.

(a) y = xy' + y' (b) yy' = x(y')2 + 1(~) (y - xy')2 + 2y' = 0 (d) y - xy' = In y'(e) y = xy' + (y')2

2. Solve the following equations by first differentiating, and then findingall values of y'. (The solution for y' is frequently simpler if one substitutes ufor y', and u' for y" after differentiation.)

(a) 3y = xy' + 2x (b) y = x + y' + xy'(c) y = xy' - In Ixl (d) y = (x + l)y' + x2

(e) y' = 1 + x - y (f) y + xy' = 2x + y'(g) 2yy' = 3x + x(y')2 (h) y = y' - x + In Iy'l3. (a) Show that the differential equation of the family y = eIn Ixl + g(e) is

y = xy'ln Ixl + g(xy').(b) Show, as in the proof of Theorem 1, that the only solutions of the

differential equation of part (a) are the functions y = e In Ixl + g(e)and the envelopes of this family.

4. Solve the following equations (see Problem 3). Show that the envelopes ofthe families of solutions obtained by inspection-can also be found by differentiating the equation as in Problem 2.

(a) y = xy' In Ixl - (xy')2 (b) y = xy' In Ixl + 1 + 1/xy'

ANSWERS

1. (a) y = ex + e(b) y = ex + lie, y2 = 4x(c) (y - ex)2 + 2e = 0, y = 1/2x(d) y = ex + In e, y = -1 - In (-x)(e) y = ex + e2, y = -ix2

2. (a) y = x + ex3

(b) y = x + e(I + x) - (x + 1) In 11 + xl(c) y = ex - 1 - In Ixl(d) y = -x2 - 2x + 2(x + 1) In Ix + 11 + e(x + 1)(e) y = x + ee-'"

e(f) y = I_x+ x + I

(g) 2y = ex2+ 31e, y = ±V3 x(h) y = ee'" + In lei, y = -(1 + x)

4. (a) y = e In Ixl - e2, y = i (In Ixl)2(b) y = cln Ixl + lie, y = 2 (In Ixl)1/2 + 1

CHAPTER 2

SPECIAL METHODS FOR FIRST ORDER EQUATIONS

2-1 Homogeneous equations-substitutions. We can change the formof a differential equation in the following way. If we let y be any SOhl

tion ofy' = F(x, y)

and introduce a new (unknown) function u by an equation such as y(x)g(u(x»), then u must satisfy the new equation

g'(u)u' = F(x, g(u»).

(1)

(2)

Conversely, if u satisfies (2), then y satisfies (1). This process of substitution is of course pointless Unless the resulting equation is simpler than theoriginal. We study here one type of equation for which a standard substitution, y = xu, always gives an equation in which the variables separate.

A function F of two variables is homogeneous (of degree zero) if and onlyif F(tx, ty) = F(x, y) for all numbers x, y, t. The following are examplesof homogeneous functions:

x+y--,x

xy + x2 sin (y/x)y2 (3)

If F is a homogeneous function, then F(x, y) depends only on the ratio y/x, .and hence F can be considered as a function of a single variable u = y/x.To see this, put t = 1/x in the definition to obtain

F(x, y) = F Gx, ~ y) = F (1, ~) = F(1, u).

The homogeneous functions (3), for example, can be written

x ~ y = 1 + (~),

x 3 + xy2 1 + (y/X)2y3 + x2y (Y/X)3 + (y/x) ,

xy + x2 sin (y/x)y2

25

(y/x) + sin (y/x)(y/x) 2

26 SPECIAL METHODS FOR FIRST ORDER EQUATIONS [CHAP. 2

A homogeneous differential equation is an equation of the form y' = F(x, y),where F is homogeneous of degree zero. If we make the substitutionu = y/x in a homogeneous equation, we get, since y' = u + xu' andF(x, y) = F(l, y/x),

u + xu' = F(l, u).

The variables separate in (4) to give

u' 1F(l, u) - u x

EXAMPLE 1., x + y 1 + (y/x)

y = x - y = 1 - (y/x) .

We let u = y/x, hence y' = u + xu', and the equation becomes

1 + uu+xu'=--·

1 - u

Simplifying, and separating variables, we get

(4)

1 + u - u + u 2

xu' = --'-----,----'---1 - u

(1 - u)u' 11 + u 2 X

Integration now yields

1 + u 2

1- u'

and hence the solutions of the original equation are given by


, x(y + 1) + (y + 1)2Y = x 2

is not homogeneous, but becomes homogeneous after the substitutionvex) = y(x) + 1. In this case v' = y', and v must satisfy

v' = xv + v2

= !!. + (!!.)2 .x2 X X

2-1] HOMOGENEOUS EQUATIONS-SUBSTITUTIONS 27

Now let u = vjx, and obtain

xu' + u = u + u 2,

-u- 1 = In Ixl + c.

Reversing the two substitutions, we get

x- v= In Ixl + c,

-x = (y + 1) [In Ixl + c].


is not homogeneous, but becomes homogeneous if we consider the independent variable to be x + 1 instead of x. If we let t = x + 1 and writedyjdt = dyjdx, we get the -homogeneous equation

Now the substitution u = y/t, dy/dt = t(dujdt) + u, yields the equation

which gives-!u-2 = In ItI+ c.

Finally, the solutions of the original equation are given by

1(x + 1)2 .-"2 -y- = In Ix + 11 + c.

The mechanical process of substituting t for x + 1, and dyjdt fordyjdx = y' can be justified by making a proper substitution. We illustrate the argument for the very simple equation

y' = -y-.x + 1

If (5) is written out completely we have

y'(x) = y(x) .x + 1

(5)

(6)


Since (6) is to be an identity for a solution y, the equation can be writtenequally well as

'( 1) _ y(x - 1)Y x - - .x (7)

[For this paragraph only, "y(x - 1)" means "y of x-I," etc.] In (7)we make the substitution

vex) = y(x - 1),

and (7) becomes

v'(x) = y'(x - 1), (8)

v'(x) = vex) .x

Separating variables and integrating, we get vex) = ex, whence [by (8)]y(x - 1) = ex, and y(x) = c(x + 1). We clearly end up with the sameresult with the formalism of Example 3. Let y = t + 1, dy/dt = dy/dx,in (5) to get

dy Ydt =-.

Integration gives y = et and, by substitution, Y = e(x + 1).The equation

, ax+by+cy = px+qy+r' (9)

with aq - bp ~ 0, can be made homogeneous by a substitution of theform

vex) = y(x + a) - {3. (10)

As indicated above, we can accomplish the same thing as the substitution(10) by writing

v = y - {3, t = x - a,dv dydt dx' (11)

With the agreement (11), equation (9) becomes

dvdt

at + bv + (aa + b{3 + e)pt + qv + (pa + q{3 + r)

(12)

If aq - bp ~ 0, we can find numbers a and {3 such that

aa + b{3 + e = 0,

pa + q{3 + r = 0.

2-1] HOMOGENEOUS EQUATIONS-SUBSTITUTIONS 29

For a and (3 which are solutions of these linear equations, equation (12)is homogeneous:

dvdt

at + bvpt + qv

a + b(vlt)p + q(vlt)

EXAMPLE 4.

, x + 2y - 4y = .

2x-y-3

Let t = x - a, v = y - {3, dvldt = dyldx, and the equation becomes

, t + 2v + (a + 2{3 - 4)v = .2t - v + (2a - (3 - 3)

The numbers a and {3 which satisfy

a + 2{3 = 4, 2a - {3 = 3

are a = 2, (3 = 1. The substitution t = x - 2, v = y - 1 [properly,vex) = y(x + 2) - 1] gives the equation

, t + 2v 1 + 2(vlt)V=--= .2t - v 2 - (vlt)

Now make the substitution u = vlt, v' = tu' + u, to obtain

u + tu' = 1 + 2u2-u

Hence

tu' = 1 + 2u - 2u + u2

,

2-u

(2 - u)u' 11 + u 2 = t'

Integration gives

2 tan- l u - ~ In (1 + u 2) = In ItI + c,

and the solutions are given by

2 tan-l (~) - ~ In (1 + ~:) = In ItI + c,

2 tan-l(; D- ~ In [1 + (~ ~rJ = In Ix - 2/ + c.


PROBLEMS

y

x+y3. y'

Solve the differential equations in 1 through 5.

1. y' = Y + x 2. y' = 2y - xx Y

2 24. y' = X + y 5. y' = _x_

xy x+ y

6. Show that if y' = F(x, y) is a homogeneous equation, and y = g(x) is asolution, then y = (l/a)g(ax) is a solution for any number a. Illustrate thisfact with the solutions of Problem 1. Interpret the result geometrically.

7. Find an appropriate substitution and solve (x - y)2y' = 1.8. Find a substitution which transforms y2 = 2yy' (x + 1) + 1 into a

Clairaut equation and solve it.

Solve the differential equations in 9 and 10.

9 , =2x-5y +3 10 ,= x+y+2.y 2x+4y-6 .y 2x+y-l

11. The procedure in the text for problems of the form

I ax+by+ey = px+qy+r

fails if aq - bp = O. Show that in this case ax + by = k(px + qy), and thesubstitution u = px + qy separates the variables.

12. Solve y'

ANSWERS

12. 1O(2x + y) + In /5(2x + y) + 71 = 25x + e

1. y = x (In Ixl + c) 2. x = (y - x) [In Iy - xl + c]3. x = y (In Iyl + c) 4. y2 = 2x2 (In Ixl + c)

5. In 11 _.it - (.it)21_ ~ In 1V5 + 1 + (2y/x) I = -2 In Ixl + ex x V5 V5 - 1 - (2y/x)

_ 1. Ix - y - 11 .7. y - 2 In x _ y + 1 + c

8. l = ex + c + 1

9. (x - 1)3 [4 (~ =~) - 1] [~ _ ~ + 2J= e

y+5+1 V510. In 11 - y+ 5_ (y + 5YI +~ In ~ 2 - 2

x-3 x-3 V5 y+5+~+V5x - 3 2 2

-21n Ix - 31 + c

2-2] EXACT EQUATIONS 31

2-2 Exact equations. The first order equation

M(x, y) dx + N(x, y) dy = 0 (1)

is called exact if there is a function u of two variables such thatu",(x, y) = M(x, y) and uy(x, y) = N(x, y). That is, a function u suchthat

du(x, y) = M(x, y) dx + N(x, y) dy. (2)

If there is a function u satisfying (2) for given functions M and N, thenexcept for an arbitrary additive constant, there is exactly one such function. We saw in Section 1-2 that the solutions of (1), if (1) is exact, arejust those functions y satisfying u(x, y) = c for some constant c. In thissection we find conditions on M and N which are necessary and sufficientthat (1) be exact and methods for finding the function u.

THEOREM 1 (Necessary condition for exactness). If (1) is exact andMy and N", are continuous, then My = N ",. Specifically, the equalityM y(x, y) = N ",(x, y) must hold for all (x, y) in any region on which thereis a function u satisfying the condition (2).

Proof. Suppose that u is a function satisfying (2); i.e., that '" = Mand Uy = N. The mixed second partial derivatives U",y and '" are equalwhere they are continuous, so we must have

My(x, y) = U",y(x, y) = uy",(x, y) = N",(x, y).

The necessary condition My = N", is also sufficient if some geometricrestrictions are made about the region on which the condition holds. Weshow this next for the special case in which the region under considerationis a rectangle. The problem is treated in more generality in the nextsection.

THEOREM 2 (A sufficient condition for exactness). If M, N, My and N",are continuous, and M y(x, y) = N ",(x, y) for all (x, y) in some rectangle R,

R = {(x, y): a ::; x ::; b and c ~ y ::; dl,

then there is a function u defined on R such that u",(x, y) = M(x, y) anduy(x, y) = N(x, y) for all (x, y) in R.

Proof. We will show that the following function u satisfies the conditionsof the theorem:

u(x, y) = fa'" M(t, c) dt +f N(x, t) dt. (3)

Note that u(x, y) is defined for every (x, y) in R, since the integrands in


d

y

c

r-----------,

: R (x, y) :I II II S II I

: I

L j--..

(a, c) (x, c) (b, c)

a

FIGURE 2-1

x b

(3) are continuous (functions of t) on the intervals of integration (seeFig. 2-1). We will need the following formulas from calculus:

iJ~ fa" M(t, c) dt = M(x, c),

iJiJxiY

N(x, t) dt = faY N,,(x, t) dt,

iJiJy i Y

N(x, t) dt = N(x, y).

(4)

(5)

(6)

Using (4) and (5), and the assumption that My = N" within the rectangle,we obtain* from (3)

u,,(x, y) = M(x, c) + leY N,,(x, t) dt

= M(x, c) + leY My(x, t) dt

= M(x, c) + M(x, t)I::= M(x, c) + M(x, y) - M(x, c)

= M(x, y). (7)

Since the first integral in (3) does not depend on y, we get immediatelyfrom (6)

uy(x, c) = N(x, y). (8)

* My is a function of two variables, and My(x, t) is the value of this functionat (x, t). Do not confuse My(x, t) with iJ/iJyM(x, t), which is zero.

2-2] EXACT EQUATIONS 33

The first integral in (3) is the integral of M restricted to the line L 1

(Fig. 2-1) from (a, c) to (x, c), and the second is the integral of N restricted to the line L 2 from (x, c) to (x, y). We need to know that thefunctions are continuous, and the identity My = N x holds along the broken-line path (L 1 and L 2) from (a, c) to any point (x, y) in R. If there were aregion in R where the conditions of the theorem failed (e.g., the shadedregion S of the figure), the argument above could not be used to showdu = M dx + N dy, even for points not in S.

Rather than use the formula of the theorem, it is frequently simpler todeal with indefinite integrals. Formula (3) for u shows that if we integrateN with respect to y, we obtain a function which differs from u by a function of x only [namely, the first integral on the right of (3)).

Schematically,

u(x, y) = JN(x, y) dy + cJ>(x).

To find cJ>, we use the fact that we must have

ux(x, y) = a~ !N(X, y) dy+ cJ>'(x) = M(x, y).

This gives a formula for cJ>'(x) from which cJ> can be found by integration.Similarly, we can write

u(x, y) = JM(x, y) dx + cJ>(y)

and find cJ> from the condition

aay

! M(x, y) dx + cJ>'(y) = N(x, y).

EXAMPLE 1. 2xy dx + (x 2 + 2y) dy = O.

Since (a/ay)(2xy) = 2x = (a/ax)(x 2 + 2y), the equation is exact. Hence

u(x, y) = J2xy dx + cJ>(y)

= x2y + cJ>(y).

The condition uy(x, y) = x2 + 2y gives us

x2 + cJ>'(y) = x2 + 2y, cJ>'(y) = 2y.

Therefore, u(x, y) = x2y + y2, and the solutions of the differential equation are the functions satisfying x2y + y2 = C.


u(x, y) = h'" M(t, 1) dt + ~y N(x, t) dt

= h'" 2t dt + ~y (x 2 + 2t) dt

= x 2 - .4 + x 2y + y2 - x 2 - 1

= x2y +y2 - 5.

The formula (3), with a = 2, C = 1, applied to the above problemyields directly

The functions defined by the equations x 2y + y2 = C and x 2y + y2 5 = c are of course the same..

PROBLEMS

Show that the following equations are exact and solve them.

1. y dx + (x + 1) dy = 0

2. sin y dx + (x cos y + 1) dy = 03. (2xy2 + 2) dx + 2x2y dy = 0

y 14. - dx = - dy

x2 X

5. [eX + y2 cos (xy)] dx + [sin (xy) + xy cos (xy)] dy = 0

6. (x2 + y2)(X dx + y dy) + 2 dx = 0

7. (In Iy + 11 + l) dx + (y ~ 1 + 2XY) diJ = 0

8. y' = _ 2x + yx+ 2y

9. With the hypotheses of Theorem 2, show that the integral of N along thesegment from (a, c) to (a, y) plus the integral of M along the segment from(a, y) to (x, y) gives a function v such that

dv = M dx + N dy.

Prove that v = u. [Hint: The functions u and v differ by a constant. Show thatu = vat some point and hence that the constant is zero.]

10. If du = M dx + N dy, then u is also given by

u(x, y) = 101

[xM(tx, ty) + yN(tx, ty)] dt.

Use this integral to solve the following equations:

(a) (1 + y) dx + x dy = 0(b) (y2 + 1) dx + 2xydy = 0(c) dx - sin y dy = 0

2-3] LINE INTEGRALS

ANSWERS

35

1. xy + y = e 2. x sin y + y = e3. X2y 2 + 2x = e 4. x + y = ex5. y sin (xy) + eX = e 6. (x2 + y2)2 + 8x = e7. x In Iy + 11 + xy2 = e 8. x 2 + xy + y2 = e

10. (a) x + xy = e (b) x + xy2 = c (e) x + cos y - 1 = e

2-3 Line integrals. The concept of an integral along a curve in the planehas many applications. We introduce the idea briefly here in order toclarify the treatment of exact equations.

A curve is a set of points of the form

c = {(j(t),g(t)):a ~ t ~ bl, (1)

where j and g are functions with continuous derivatives on [a, b]. We willsay that .

x = jet), y = get), (a ~ t ~ b) (2)

are parametric equations for C. The curve C in (1) is closed if (j(a), g(a)) =(j(b), g(b)) and is a simple closed curve if (j(tl), g(tl)) = (j(t 2 ), g(t2 ))

implies t l = a and t2 = b, or t 1 = band t2 = a. That is, a simpleclosed curve does not intersect itself. (See Fig. 2-2.)

Simple closed curve Nonsimple curve

FIGURE 2-2

If M and N are functions of two variables which are continuous on the__ curve C in (1), then we define the line integral of M(x, y) dx + N(x, y) dy

along C as

Ie (M(x, y) dx + N(x, y) dy)

= lab [M(j(t), g(t))f'(t) + N(j(t), g(t))g'(t)] dt. (3)


A curve C can be represented parametrically by infinitely many pairs offunctions (f, g). The integral on the fight side of (3), however, will givethe same value for any pair of functions representing C if the curve istraced out in a given direction for increasing parameter values. Thedefinition (3) therefore depends only on the set of points C, with a prescribed direction or orientation along C, and not on the particular choiceof parametric functions for C.

The line integral (3) has the following physical interpretation. If Mand N are respectively the horizontal and vertical components of a forcefield defined on the plane, then the integral (3) is the work done by thisforce field on a particle moving over the curve C in the prescribeddirection.

As special cases (N(x, y) = 0 or M(x, y) = 0) of (3), we have theformulas

fe M(x, y) dx = lab M(f(t), g(t»)f'(t) dt,

fe N(x, y) dy = lab N(j(t), g(t»)g'(t) dt.

(4)

(5)

EXAMPLE 1. The line segment C from (0, 0) to (1, 1) can be represented by the functions f(t) = t, g(t) = t, 0 ::; t ::; 1, and equally wellby the functions h(t) = t2

- 1, k(t) = t2- 1, 1 ::; t ::; 0. We com

pute f e (M dx + N dy) for the functions M(x, y) = x, N(x, y) = xy,using both the above parameterizations, to illustra,te that the integraldepends only on C.

fe (x dx + xy dy) = fo 1

[t + t2

] dt = ! + !" = i,

( (x dx + xy dy) = (,(2 [(t 2- 1)2t + (t 2

- 1) 22t] dtJe Jl .

= h,(2 (2t 5- 2t3

) dt = t(8 - 1) - !(4 - 1) = l

E:xAMPLE 2. We compute fe -y dx, where C is the unit circle orientedin the counterclockwise direction. A convenient parametric representationis x = f(t) = cos t, Y = g(t) = sin t, 0 ::; t ::; 27l". We obtain

((2" (2"Je -y dx = Jo

-sin t (-sin t) dt = Jo

sin 2 t dt = 7l".

EXAMPLE 3. We compute fex dy, where C is the triangle consisting ofthe segments C1 : (0,0) to (1, 0); C2 : (1,0) to (0, 1); and C3 : (0, 1) to (0, 0).

2-3] LINE INTEGRALS 37

The integral is the sum of. the separate integrals taken over C ll C2 , Ca.We have

r xdy = 0,lCI

or x dy = r x(-1) dx = i,lC2 11

r x dy = O.lCa

The first integral is zero since on C1 , y is constant, and dy = 0 (that is,the function g in (5) is constant, and g'(t) == 0). The third integral is zerosince x = N(x, y) == 0 on Ca. For C2 we used x as the parameter, withy = -x + 1, dy = (-1) dx, and x running from 1 to 0 to match theorientation of C2 • A more formal procedure would be x = f(t) = 1 - t,Y = get) = t, 0 :::; t :::; 1, giving

f 11 21·x dy = (1 - t) dt = [t - it 10 = t.

C2 0

It is not accidental (cf. Problem 7) that the line integrals of Examples 2and 3 give the areas bounded by the closed curves.

A domain is a set G of points in the plane such that (i) G is open-eachpoint of G is the center of a disc contained in G, and (ii) Gis connectedany two points of G can be joined by a curve lying in G. The conditionthat G be open means that the set G does not contain any of its boundarypoints. For example, the set of points inside a simple closed curve, excluding the points of the curve itself, is a domain.

The next theorem shows that if M dx + N dy is an exact differential,then the function u such that du = M dx + N dy can be found as a lineintegral of M dx + N dy.

THEOREM 1. If M(x, y) dx + N(x, y) dy = du(x, y) for all (x, y) in adomain G, then

Ic (M(x, y) dx + N(x, y) dy) = u(c, d) - u(a, b)

for every curve C in G from (a, b) to (c, d).

Proof. Let C be any curve in G from (a, b) to (c, d), and let

x = f(t), y = get),

be parametric equations for C. In particular, (j(a), g(a»)(j(f3), g(f3») = (c, d). Define a function H on [a, f31 by

H(t) = u(j(t), get»).

(a, b), and


By our assumption that U x = M, U y = N, we have

H'(t) = ux(j(t), g(t))f'(t) + uy(j(t), g(t))g'(t)

= M(j(t), g(t))f'(t) + N(j(t), g(t))g'(t).

It follows that [cf. (3)]

fe (M(x, y) dx+ N(x, y) dy) = laP H'(t) dt

= H({3) - H(a)

= u(j({3), g({3)) - u(j(a), g(a))

= u(c, d) - u(a, b).

For expressions M dx + N dy which are exact differentials, we will write

j (C,d) ( )M(x, y) dx + N(x, y) dy

(a,b)

to indicate the line integral taken over any curve from (a, b) to (c, d). Byfixing a point (a, b), and taking any point (x, y) for (c, d) in (6), we get thefollowing representation for a function u such that du = M dx + N dy.

j (x,y) ( )u(x, y) = M(x, y) dx + N(x, y) dy .

(a,b)(6)

The formula given for u in Section 2-2 [Formula (3)] is just the lineintegral along a broken line path from (a, c) to (x, y).

We return ~ow to the problem of determining to what extent the condition My = N x is sufficient that M dx + N dy be exact. The relevanttheorem depends on the following geometric notion. A domain G issimply connected if the region bounded by any simple closed curve in Gcontains only points of G. This says, roughly speaking, that G has noholes in it; i,e., no curve in G can surround points which are not in G.The following theorem is a conseqtlence of Green's theorem (see Problem 6),and we will omit the proof.

THEOREM 2. If My(x, y) = Nx(x, y) for all (x, y) in a simply connecteddomain G, then there is some function u defined on G such that du(x, y)M(x, y) dx + N(x, y) dy for all (x, y) in G.

Once we know that there is a function u defined on G, the function canbe determined by the integral (6). The effect of picking different initialpoints (a, b) is simply to change the function u by an additive constant.The statements of Theorem 2, Section 2-2, are restricted to points within

2-3] LINE INTEGRALS 39

a rectangle. A rectangle is merely an example of a simply connected set,and moreover one in which every point can be reached with the simplebroken line path used in that theorem. The formula of Problem 10,Section 2-2, is the line integral over the line segment from (0,0) to (x, Y).

PROBLEMS

1. Compute the integrals for the curves specified.

(a) fe [(x + 1) dx - xy2 dy]; C is the parabola y = x2 from (-1, 1)to (2,4).

(b) fe [(x2 + y) dx + (x + y) dy]; C is the closed curve consisting of thearc of y2 = x from (1,1) to (1, -1), and the segment from (1, -1)to (1,1).

(c) fe [x2y dx + Ij(x2 + 1) dy]; C is the boundary of the unit square{(x, y) : 0 :$; x :$; 1, 0 :$; Y ~ I} oriented counterclockwise.

2. Calculate fe(x + y) dx for the two curves Cland C2. CI is the segmentfrom (0,0) to (1, 1), and C2 is the broken line from (0,0) to (1,0) and (1,0)to (1, 1). Conclude that the hypothesis of Theorem 1 is necessary for a lineintegral to be independent of path.

3. Calculate fial (2xy dx + x2 dy), where the integral may be computedalong any convenient curve.

4. (a) Is the domain Go consisting of all points of the plane except (0, 0)simply connected?

(b) Is the domain G consisting of all points of the plane except those ofthe form (x, 0), x :$; 0, simply connected? .

5. Verify that the exactness condition of Theorem 2 holds for

-y dx + x dyx2 + y2 x2 + y2

except at (0,0). Calculate

1 -y X

2 + 2 dx + 2 + 2 dyex y x Y

for CI: the upper half of the unit circle from (1, 0) to (-1, 0) and for C2: thelower half circle from (1,0) to (-1,0). Reconcile this with Theorems 1 and 2.

6. Green's theorem states that if M, N, My, and N", are continuous on theset G consisting of a simple closed curve· C and its interior, and if C is orientedin the counterclockwise direction, then

f£ [N",(x, y) - My(x, y)] dy dx = fe M(x, y) dx + N(x, y) dy.

Prove this for the case N(x, y) == 0, when C is the closed curve formed byy = (I(x) and y = f(x), with (I(a) = f(a), (I(b) = feb), and (I(x) < f(x) fora < x < b.


7. Use Green's theorem (Problem 6) to show that fe-Y dx and fex dy givethe area bounded by the simple closed counterclockwise curve C.

8. Show that x = a cos t, y = b sin t, 0 ~ t ~ 211", are parametric equations of the ellipse x2ja2 + y2jb2 = 1. Use either formula of Problem 7 tofind the area bounded by this ellipse.

9. Solve the differential equations and specify the domains on which the solutions hold.

I I(a) dx - dy = 0vx=y vx-y

(-1 Y xy ) x

2

(c) tan - - 2 + 2 dx + 2 + 2 dy = 0x x y x Y

10. The differential of Problem 5 is exact in the simply connected domainbounded by the simple closed curve of Fig. 2-2. There is, therefore, a function udefined on this domain such that u(l, 0) = 0 and du(x, y) = -yj(x2+ y2) dx+xj(x2+ y2) dy. Find u(-I, 0) and u(-2, 0).

ANSWERS

1. (a) _415l 2. I, 2 3. x2y - 2

(b) 0(c) -f

4. No, yes 5. 11", -11" 8. 1I"ab

9. (a) x - y c2 for points below y = x(b) x2 + yIn Ixl = c for the right half-plane, or the left half-plane(c) x tan-1 (yjx) = c for the right half-plane, or the left half-plane

10. u(-I,O) = -11", u(-2,0) = 11"

2-4 First order linear equations - integrating factors. A linear equationis any equation of the form

y(n) + Pn_1(X)y(n-1) + ... + P1(X)Y' + Po(x)y = q(x).

We will study linear equations in detail in Chapters 3 and 4. Here wetreat the simplest case, the first order linear equation

y' + p(x)y = q(x). (1)

We assume that the functions P and q are continuous on some interval[a, b]. With this assumption it follows from Theorem I, Section 1-4, thatthere is a solution of (1) through each point (xo, Yo) such that a < Xo < b.Although we cannot conclude from this theorem that the solutions of (1)are defined on the whole interval [a, b], this is the case, as we show below.

2-4] FIRST ORDER LINEAR EQUATIONS 41

First consider the special case of (1) in which q(x) == 0,

y' + p(x)y = O. (2)

In (2) we can separate the variables and solve as follows;

y = 0;or

y = 0;or-p(x),y'-=y

In Iyl = - J p(x) dx + In c (c > 0),

Iyl = ce-fp(X) dx (c > 0), or y = 0;

y = ce-fp(X) dx (c arbitrary).

If we write the solutions in the form

yefP(X) dx = c,

then differentiation gives

y'efP(x)dx + p(x)yefp(x)dx = o.

That is, Eq. (2) becomes exact if multiplied by

fp(x)dxe . (3)

If a differential equation becomes exact after multiplication by somefunction, such as (3), this function is called an integrating factor for theequation.

Now return to equation (1), and multiply both sides of this equationby the function (3). We get the equivalent equation

y'efP(x)dx + p(x)yefP(x)dx = efP(x)dxq(x). (4)

The left side of (4) is an exact differential, as we have seen. Since the rightside of (4) does not involve y, it also is an exact differential, and thereforeefP(x)dx is an integrating factor for (1) as well as (2). The solutions of (4),and hence (1), can be written

yefP(x)dx = JefP(X)dXq(x) dx + c,

or, in explicit form,

y = e-fp(x)dx [JefP(x)dxq(x) dx + cJ. (5)


We assemble these facts in the following statement.

. THEOREM 1. If p and q are continuous on [a, b], and PI(x) = p(x) forall x in [a, b], then eP(x) is an integrating factor for (1). The solutionsof (1) are defined on all of [a, b], and are the functions

y(x) = e-P(X) Lt eP(t)q(t) dt + clEXAMPLE 1. y' + 2xy = 3x.

We can write the solutions directly from Formula (6).

y = e-f2xdx [!ef2XdX3x dx + c]= e-

x2 [fex2

3x dx + c]2 2= e-x [feX + c]

.a + _x2

= 2 ce .

Alternatively, we find the integrating factor ef2x dx = ex2 and write

eX2 yl + yeX22x = ex23x.

Integration gives

EXAMPLE 2.1 2

y' - - y = .x - x 2

The integrating factor is

Multiplying by l/x, we get

1 I 1xy - x 2 Y =

and hence the solutions are given by

2--,x 3

1 1-y=-+cX x2 '

1 cy = - +-.

x x


N ow we look at an example of an integrating factor for a nonlinearequation.

EXAMPLE 3. Y dx + (2xy2 + x) dy = O.

This equation is not exact (My = 1 ;= N x = 2y 2 + 1) and is not linear,since y2 appears. We write the equation iIi the form

x dy + y dx + 2xy2dy = O.

The part x dy + y dx = d(xy) will remain exact if multiplied by anyfunction of xy. The remaining term, 2xy2dy, becomes an exact differentialif multiplied by (xy)-l, which is therefore an integrating factor. Theequation becomes

(xy)-I(X dy + y dx) + 2y dy = 0,

ord [In Ixyl + y2] = O. (6)

Notice that y = 0 is a solution of the original equation, but not of (6).The solutions therefore satisfy

y = 0, or In Ixyl + y2 = C,

which can be expressed in the single formula

Finding integrating factors is gE:meraJly just a matter of recognizing thecommon differentials. However, we can determine whether a first orderequation has an integrating factor which depends only on x and give aformula for the integrating factor in this case.

Suppose the equation

M(x, y) dx + N(x, y) dy = 0 (7)

has the integrating factor jl(x); that is, suppose the following equation isexact.

jl(x)M(x, y) dx + jl(x)N(x, y) dy = O.

- By the results of Section 2-2, we must have

a aay [jl(x)M(x, y)] = ax [jl(x)N(x, y)],

orjl(x)My(x, y) = jl'(x)N(x, y) + jl(x)Nx(x, y).

(8)


Rearranging this, we get

J.L'(x) My(x, y) - Nx(x, y)J.L(x) = N(x, y) (9)

Hence (9) is a necessary and sufficient condition that (8) be exact. Sincethe left side of (9) depends only on x, it follows that the right side of (9)must be a function of x only.

Hence !J. must satisfy

My(x, y) - Nx(x, y)N(x, y)

q,(x) .

from which we get

J.L'(x)J.L(x)

q,(x),

We have proved the following theorem.

THEOREM 2. If [My(x, y) - Nx(x, y)]jN(x, y) = q,(x) , a function ofx only, then e/</>(x) dx is an integrating factor for Eq. (7).

EXAMPLE 4. (y2 + 1 + x) dx + 2y dy = O.

The condition of Theorem 2 is satisfied, since

2y - 02y

1,

which does not involve y. Therefore e/ldx = eX is an integrating factor,and

is exact. The solutions are

PROBLEMS

1. Solve the following equations:

(a) y' + (ljx)y = 4x2

(c) y' + 2xy = 2x3

(e) y' + (cot x)y = 3 sin x cos x

(b) y' + y = e"(d) (x - l)y' + xy = e-"(f) y" + 2y' = 4x


2. Show that l/x2, l/y2, and 1/(x2 + y2) are integrating factors forx dy - y dx = O.

3. Find an integrating factor and solve.

(a) 2y dx + x dy = 0(b) 2xy2 dx + 3x2y dy = 0(c) (x - y) dx + x dy = 0(d) (2y3 + 6xy2) dx + (3xy2 + 4x2y) dy 0(e) (x in Ixyl + x) dx + (x2/y) dy = 0(f) (-vi" + y) dx + 2x dy = 0

4. An equation of the form

y' + p(x)y = q(x)yn,

or (m = -n),ym y' + p(x)ym+l = q(x),

is called a Bernoulli equation. Find a substitution which transforms a Bernoulliequation into a linear equation in a new unknown.

5. Solve the following equations:

(a) yy' - (2/X)y2 = 1(b) y' + y = 3exy3

6. Show directly from Theorem 1, Section 1-4, that if y is a solution of (2)and y(xo) = 0 for some xo, then Vex) == o.

7. Assume Yo is a nonzero solution of (2), and Yi is a solution of (1).

(a) Show by substitution that Yi + eyo is a solution of (1) for everynumber e.

(b) Show by Theorem 1, Section 1-4, that every solution y of (1) can bewritten y = y~ + eyo for some e.

8. Assume that Yi and Y2 are distinct solutions of (1), and let yO = Yi - Y2.

(a) Show that Yo is a solution of (2).(b) Write all solutions of (1) in terms of Yi and Y2 (cf. Problem 7).

9. Check that the functions below are solutions of the given linear equation,and write all solutions of each equation by the method of Problem 8.

(a) 1 - e-x , 1 + 2e-x ; y' + y = 1(b) x - l/x, x + l/x; y' + (l/x)y = 2(c) x, x2; (x2 - X)y' + (1 - 2x)y = -x2

10. Show that if Yo is a nonzero solution of (2), then (1) can be solved by thesubstitution Vex) = yo(x)v(x), and the solutions are y = Yo[f(q/yo) dx + e].(This method, called "variation of parameters," extends to higher order linearequations.)

11. Solve the equations of Problem 9 by the variation of parameters method(Problem 10).


ANSWERS

3. (a)x 2y = c(b) x2y3 = c(c) V/x + In Ixl = c(d) x 2y3 + 2x3y2 = c(e) xIn Ixyl = c(f) x + 2yVx = c

8. (b) Y = YI + C(YI - Y2)

1. (a) Y = x3 + c/x(b) y = tex + ce-X

(c) y = x2 - 1 + ce-x2

(d) eX(x - l)y = x + c(e) y = sin2 x + c esc xCO V = x2 - X + cIe-2x + C2

5. (a) y2 = -~x + cx4

(b) y2(6eX + ce2x) = 1, Y = 0

9. (a) y = 1 + ce-X

(b) y = x + c/x(c) y = x + c(x2 - x)

y' = F(x, y)

2-5 Orthogonal families. We will say that two curves are orthogonal ifand only if their tangents at points of intersection are perpendicular lines.Two families of curves are orthogonal if and. only if every curve of thefirst family is orthogonal to every curve of the second family, and viceversa. Given a family of curves, the orthogonal fam~'ly, or family of orthogonal trajectories, is the set of all curves which are orthogonal to everycurve in the given family.

Orthogonal· families of curves occur frequently in physical situations.For example, the lines of force in a two-dimensional force field are orthogonal to the equipotential curves. In the study of heat conduction, onefinds that the isothermal curves are the orthogonal trajectories of the linesof heat flow.

If

,_ 1y - - F(x, y)'

is the differential equation of a family of curves, then the line tangent toone of these curves at a point (x, y) has slope F(x, V). For an orthogonaltrajectory of the family, the tangent line at (x, y) must have slope-l/F(x, V), since lines are perpendicular if and only if their slopes arenegative reciprocals. Hence the differential equation of the family oforthogonal trajectories is

There are two steps in finding the family orthogonal to a given oneparameter family of curves. First, find the differential equation of thefamily by differentiating and eliminating the constant. We repeat:eliminate the constant. Second, replace y' by -l/y' to obtain the differential equation of the orthogonal family and solve this differential equation.

EXAMPLE 1. Find the family orthogonal to y = cx2•

2-5] ORTHOGONAL FAMILIES

FIGURE 2-3

47

y = txy'.

or

Differentiation gives y' = 2cx, or c = y' j2x. Substituting this valueof c in the original equation, we obtain the differential equation of thefamily,

The differential equation of the orthogonal family is therefore

y = ~x (- ~,),

4yy' + 2x = O.

The orthogonal family is the family of ellipses (see Fig. 2-3),

2y2 + x 2 = c2.

EXAMPLE 2. Show that the family y2 = 2cx + c2 is self-orthogonal~

Differentiating, we get 2yy' = 2c, or c = yy'. Therefore, the differentialequation of this family is

y2 = 2xyy' + (yy') 2.

Substituting -ljy' for y' we obtain the differential equation of theorthogonal family

or

Since this last equation is the same as that for the given family, thefamily is self-orthogonal.

48 SPECIAL METHODS FOR FIRST ORDER EQUATIONS

PROBLEMS

[CHAP. 2

Find the orthogonal families and sketch both families in Problems 1 through 5.

2. y = ex4

5. y = ee"

6. Sketch the self-orthogonal family of curves of Example 2.7. Show that if every curve in the family of rectangular hyperbolas xy = e

is rotated 45°, the resulting family is orthogonal to the original.8. Show that x2je + y2j(e - a 2) = 1 is a family of ellipses and hyperbolas

with foci at (±a, 0). Show that the differential equation of the family is(x + yy')(x - yjy') = a 2, and hence that the family is self-orth~gonal.

ANSWERS

1. y = ex 2. x2+ 4y2 = e2

4. y2 = In Ixl + tx2+ C

3. x2+ (y - e)2 = e2

5. y2 = -2x + e

2-6 Review of power series. An indicated infinite sum of the form

<Xl

L anxn = ao + alx + a2x2 + ... ,n=O

or of the form

(1)

<Xl

L an(x - c)n = ao + al(x - c) + a2(x - C)2 + ... (2)n=O

is called a power series. We recall a few of the basic facts about powerseries.

The series (1) converges at Xo to f(xo) if and only if

(3)

If the limit (3) fails to exist, the series (1) diverges at Xo. If (1) convergesat Xo and IXll < Ixol, then (1) converges at Xl. If (1) diverges at Xo andIXll > Ixol, then (1) diverges at Xl' It follows that (1) converges only atzero, or converges everywhere, or there is a positive number r such that(1) converges on (-r, r) and diverges outside [-r, rl. The number r iscalled the radius of convergence, and the interval (-r, r) is called theinterval of convergence of (1). A series mayor may not converge at eitherendpoint of its interval of convergence. For a series of the form (2), theinterval of convergence is an interval of the form (c - r, c + r).

2-6] REVIEW OF POWER SERIES 49

Power series converge absolutely on their intervals of convergence; that is

and00

L: lan(x - ctln=O

converge on the intervals of convergence of

and00

L: an(x - c)n.n=O

This insures that we can rearrange the terms of a power series withoutaffecting the convergence, or the value of the sum. Another consequenceof absolute convergence is that one can multiply two power series bywriting down in any order all products containing one term from eachseries.

Power series converge uniformly on any closed interval in the interval ofconvergence, but not in general on the whole interval of convergence.That is, if (-r, r) is the interval' of convergence of

00

L: anxn = f(x),n=O

and -r < a < b < r, then for every E > 0, there is some No not depending on x, such that

for all x in [a, b), whenever N 2:: No. In other words, the polynomialwhich is the Nth partial sum of the series will be uniformly close to f(x)on [a, b), provided N is sufficiently large. If the series converges on thewhole. line, then it converges uniformly on any finite interval [a, b).

If f(x) = L~=O anxn for x in the interval of convergence, (-r, r), thenf is continuous and has derivatives of all orders on (-r, r). The derivativesof f are given by the series obtained by termwise differentiation of theseries for f. That is, termwise differentiation does not change the intervalof convergence of a series, and if

00

f(x) = L: anxn, x in (-r, r), (4)

n=Othen

00

f'(x) L: n-l xin(-r,r). (5)nanx ,n=l


A power series may be integrated term by term over any closed intervalwithin the interval of convergence. Thus if f is given by (4), r is theradius of convergence of the series, and -r < a < b < r, then f~f(x) dxexists since f is continuous on [a, b]. Moreover, the series (of constants)

will converge, and the following identity holds:

By integrating over the interval [0, x], or [x, 0], for Ixl < r, we can obtainindefinite integrals (antiderivatives) of 1:

-r < x < r.

The series on the right above, obtained by termwise integration of the. series for f, will have the same radius of convergence as the series for f.

We will say that f is (real) analytic on an interval (a, b) if there is a.series of the form (2) which converges to f on some interval around c, foreach c in (a, b). Any function given by a convergent power series is analyticon the interval of convergence of th{) series. Thus if f(x) = L:anxn forIxl < r, and lei < ~,there is a series L:bn(x - c)n which converges tof(x)on some interval around c.

The common functions eX, sin x, and cos x are analytic on the wholeline; their series at zero converge for all x:

sin x = t (_I)n+l 1 x 2n-\n=l (2n - 1)1

. f... ( l)n 1 2ncos X = L..J - (2 ') x .

n=O n.

(6)

(7)

(8)

The function vx is analytic on (0, (0). If c > 0, there is a series for vxof the form L:~=o an(x - c)n which will converge on (0,2c). The function1/(1 + x 2

) is analytic on the whole line. Thp series for 1/(1 + x 2) at zero

is the geometric series 1 - x 2 + x 4 - x 6 + ... which converges on

2-6] REVIEW OF POWER SERIES 51

(-1, 1). The series at the point c will have radius of convergence VI + c2

(the distance between the complex numbers c and i = v=I).The simplest way of determining the radius of convergence is the ratio

test. If

lim I~I- rn-+co a,,+l - ,

then r is the radius of convergence of (1) or (2). If

1· lan+l! 0Im--=,n~oo an

(9)

(10)

[note that the ratio in (10) is the reciprocal of that in (9)], then the seriesconverges for all x.

If the ratio test fails because Ianian+II fails to converge, one canfrequently use the following comparison test. If lanl :s; Ibnl for all n, and:Ebnxn has radius of convergence r, then :Eanxn has radius of convergence at least r. For example, the ratio test fails for the sine series (7)since every other coefficient is zero, but the series can be shown to convergeeverywhere by comparison with the exponential series (6).

If a function j is analytic on an interval around c, then the coefficientsin the series for j at care .

1a = - j(n)(c)n n! '

and hence the series is

n = 0, 1, 2, ... , (11)

PROBLEMS

1. Find the interval of convergence for

2. Find the series and its interval of convergence for the indicated functionabout the given point.

1(a) 1 _ x' c = 0 (b) VX, c = 1

(c) In x, c = 1 (d) In lxi, c = -1


3. Use the geometric series Problem 2(a) to find series for these functions.Give the interval of convergence in each case.

1 1(a) 1 _ x 2 (b) 1 + 2x (c) In (1 + 2x)

. 2x 1 I(d) (1 _ x 2)2 (e) 2 + x (f) tan- x

4. Use (6), (7), and (8) to find series for

(a) ex2 (b) sin (2x) (c) cos vi; (d) sin (x - 1).

5. Derive formula (11). [Hint: If f(x) is given by (4), then f(O) = ao. Sincej'(x) is given by (5),1'(0) = aI, etc.]

6. Suppose Lanxn converges if and only if -1 ::; x < 1. For what valuesof x do the following converge?

ANSWERS

1. (a) (-1,1) (b) x = 0 (c) (-2,2) (d) (-1,1)

00

(b) L: (-"-2txn

,

n=O

00

2. (a) L: xn

, (-1,1)n=O

(b) 1 + t (_I)n+I 1· 3·5 . ',~~2n - 1) (x _ I)n,n=I n.

00 ( )n+I(c) L: -In (x - I)n, (0,2)

n=I

(d) t -1 (x+ l)n, (-2,0)n;"1 n

3. (a) t x2n

, (-1,1)n=O

(0,2)

(c) t ~~): xn+I, (-t, t) (d) t 2nx2n

-I, (-1,1)

n=O n=O

00 (_I)n n (f) ~ (-It 2n+I (-1 1)(e) L: 2n+I x, (-2,2) L.J2n+l x "

n=O n=O00 2n 00 22n-I

4. (a) L: xnl

(b) L: (-It+ Ix

2n-

I

n=O " n=I (2n - I)!

(c) L:noo=o (-It (21n)! x

n(d) t (-It+

I1 (x - 1)2n-I

n=I (2n - I)!

6. (a) 1 ::; x < 3 (b) -4 < x ::; -2 (c) -1 < x < 1

2-7] SERIES SOLUTIONS 53

2-7 Series solutions. Even the simplest type of differential equationmay have solutions which cannot be written in terms of elementary functions, such as polynomials, exponentials, trigonometric functions, etc.For example, the solutions of

(1)

are simply the antiderivatives of e-:z:2

, but we cannot express these with asimple formula. We can, however, write a power series for each solutionof (1). Since

:z: + 1 2+ + 1 n+e=l+x -x ... -x ...2! n!

for all x, it follows that

e-:z:2= 1 _ x 2 + ~ x4 _ ••• + -!.- (_x 2 )n + . . . (2)

2! n! '

and hence the solutions of (1) are given by

1:z:

_~ 3 1 5c+ e dt=c+x-tx +--x - ....° 5· 2!

In general we do not know whether the solutions of a given differentialequation are analytic, so we proceed as follows. We suppose that a differential equation has a series solution,

(3)

and substitute the series in the equation to determine what the coefficientsmust be. That is, we find conditions on ao, at, a2, ... , which are necessaryfor (3) to be a solution of the given equation. Then we must check whetherthe series found in this way converges, and whether it is in fact a solution.

EXAMPLE 1. Suppose that

y' + 2xy = 1 (4)

has a solution y which is analytic on some interval around zero. Thus

(5)

We know that y' is given on the same interval by the series


Therefore (5) is a solution of (4) if and only if

al + 2a2X + ... + nanXn- 1 + ...+ 2x(ao + alX + ... + nanXn + ...) = 1 (7)

(8)

is an identity on some interval around zero. Collecting terms, we see that(7) is equivalent to '

al + (2ao + 2a2)X + (2al + 3a2)X2 + ...+ (2an_2 + nan)Xn- 1 + ... = 1.

Since two power series converge to the same function if and only if thecorresponding coefficients are equal [by (ll), Section 2-6), and the rightside of (8) can be regarded as the series with coefficients 1, 0, 0, 0, ... ,it follows that (8) is equivalent to the relations

al == 1,

2ao + 2a2 = 0,

2al + 3aa = 0, (9)

2an_2 + nan =,0.

The recursion relation

(n ~ 2), (10)

determines an, for n\ odd, in terms of al = 1. Thus aa = -i, as = ls,and in general "., . -

(-lt2n

a2n+l = (2n + 1)(2n - 1) ... 5·3

(-1)n2n(2 . 1)(2 . 2) ..• (2· n)(2n + I)!

(_I)n2 2nn!

(2n+l)!

Similarly, (10) determines an,fo~ in terms of ao, which is arbitraryWe find that . (' n .

-1) ao ( )a2n = n! ' n ~ 1 .

Hence y is an analytic solution of (4) if and only if y is given by

_ ~ (_I)n 2n +~ (_I)n2 2nn! 2n+ly - ao L..J n! x L..J (2n + I)! x .

n=O n=O

(ll)


The ratio test shows that both series in (11) converge for all x. Thereforeour thus far formal manipulations are justified, and there are analyticsolutions-of (4). The treatment of Section 9 would give us the solutions (11)in the form

EXAMPLE 2. Find the series in (x - 1) for the solutions of

y' = y + x 2.

Suppose that (12) has an analytic solution

y = ao + al(x - 1) + a2(x - 1)2 + ....The function x 2 has the following series in powers of (x - 1):

(12)

(13)

Hence (13) is a solution of (12) if we have the identity

al + 2a2(X - 1) + 3a3(X - 1)2 + ...= ao + al(x - 1) + a2(x - 1)2 + ... + 1 + 2(x - 1) + (x - 1)2

(ao + 1) + (al + 2)(x - 1)

+ (a2 + l)(x - 1)2 + a3(x - 1)3 + ....The coefficients must satisfy the relations

al = ao + 1,

2a2 = al + 2, a2 = !(ao + 3),

1a3 = 3! (ao + 5),

1a4 = 4! (ao + 5),

1an = ,- (ao + 5), ·(n :2: 3).n.

Conversely, if the series with the coefficients determined above converges,then it is a solution. The series is

y = ao + (ao +l)(x - 1) + !(ao + 3)(x - 1)2

'" 1+ L ,- (ao + 5)(x - l)n, (14)n=3 n.

which converges for all values of x.


The method of the preceding examples consists simply of substitutingan arbitrary series in the differential equation and determining the coefficients. If only a few terms of the series are wanted, as an approximationto the solution, it is frequently easier to determine the coefficients directlyfrom formula (11), Section 2-6.

EXAMPLE 3. Find the first five terms of the series in (x - 1) for thesolution y of

y' = x + y2, (15)

such that y(I) = 1.Assuming that the solution of (15) such that y(I) 1 is analytic around

1, the solution must be

y = t -\ y(n)(I)(x - I)n.n=O n.

The initial condition gives the first coefficient, ao = y(I) = 1. The secondcoefficient, al = y'(I), can be determined from (15).

y'(l) = 1 + [y(I)]2 = 1 + 1 = 2.

By differentiating (15), we get

y' = x + y2,

y" = 1 + 2yy',

y'" = 2(y')2 + 2yy",

y(iv) ---.: 4y'y" + 2y'y" + 2yy"',

Hence the first five terms of the series are

y'(I) = 2,

y"(I) = 5,

y"'(I) = 18,y(iv) (1) = 96.

y = 1 + 2(x - 1) + %(x - 1)2 + 3(x - 1)3 + 4(x - 1)4 + ....

PROBLEMS

1. Find the series in x for the solutions of y' = y.2. Find the series in x for the solution of y' = y + x, yeO) = -1.3. Use (14) to 'show that the solution of (12) such that y(l) = 0 is

y = 5e(%-I) - 5 - 4(x - 1) - (x - 1)2.4. Find the series in x for the solutions of y' + xy = x. Show that the

solutions are

00 1 ( x2)n _%2/2

y = 1 + (ao - 1) ?; n! - 2 = 1 + (ao - l)e .

5. Find the series in x for the solutions of (1 - x)y' + y = 2x. What is theinterval of convergence of the series?


6. (a) Find all series in (x - 2) for solutions of (x - 2)y' + y = 2x - 1.(b) Find all solutions, and explain why only one series is found in part (a).

7. Find the series in x for the solutions of y' + y = (x + 1)2.8. Use (11), Section 2-6, to find the series in (x - 1) for the solution of

y' = x 2 - y such that y(I) = O. Show that the series equals _e-(x-l) +(x - 1)2 + 1.

9. Use (11), Section 2-6, to find the first five terms of the series in x for thesolution of y' = x + y2 such that yeO) = 1.

10. Use (11), Section 2-6, to find the first three nonzero terms of the solutionof y' = 1 + y2, yeO) = O. Show that the solution is y = tan x and check theresult by evaluating the first five derivatives of tan x at O.

11. Find the series in x for the solutions of y" + y = 0, and show [cf. (7)and (8) of Section 2-6] that they are y = ao cos x + al sin x.

ANSWERS

00 11. y = ao L - x

n

n=O n!

2. y = -1 - x00 2 n

5. y = ao(I - x) + L n(n _ 1) x, -1 < x < 1n=2

6. y = (x 2 - X + c)j(x - 2). The only solution analytic around 2 isy = 3 + (x - 2).

. + + 1 + ) 2 ~ n+l 1 n7.y ao (I-ao)x 2(1 aox+(I-ao)~(-I) nIx

)+ 1 )2 +~ )n+l 1 n8. y (x - 1 -(x - 1 LJ (-1 - (x - 1)2 n=3 n!

3 2 8 3 34 49. y 1 + x + 2! x + 3! x + 4! x + ...

2 3 16 510. y = x + 3! x + 5! x + ...

11. a2n = (_I)n (2:)! ao, a2n+l (_l)n (2n ~ I)! al

CHAPTER 3

LINEAR EQUATIONS

3-1 Introduction. A linear equation is one of the form

y(n) + Pn_l(X)y(n-O + ... + Pl(X)Y' + PO(X)y = q(x). (1)

We will always assume that Po, Pb ... , q are all defined and continuouson some interval (a, b). If the right-hand member, q, in (1) is zero, wesay (1) is a homogeneous linear equation. The following are some examplesof linear equations:

y' + xy = eX,

y(iV)'+ 3y' + eXy = 0,

O"+!!:.-O,+q,O= 0m 1 '

ks"--s'=g

m '

y" = eX,

y" + eXy' + x2 y = sinx.

The class of linear equations is large enough to encompass a great manyof the most useful and frequently encountered differential equations. Atthe same time, the linear equation is sufficiently specialized to admit avery comprehensive and elegant theory. We will develop this theory indetail and see that it provides simple methods of solving a large class ofof linear equations. Our principal tool will be the basic existence anduniqueness theorem for linear equations. As we indicated in Section 1-4,there is a general existence and uniqueness theorem for nth order equationsof the form

y(n) = F(x, y, y', ... ,y(n-O). (2)

However, Eq. (1) is much more restrictive than (2), and we should expecta better theorem for linear equations because of the stronger hypothesesinherent in the form of (1). Here is the theorem (to be proved inChapter 7) for .linear equations.

THEOREM 1. If Po, PI, ... , Pn-b q are continuous on (a, b), and Xo isany number in the interval (a, b), and Yo, y~, ... , y(ij-l) are any numbers,

58

3-2] TWO THEOREMS ON LINEAR ALGEBRAIC EQUATIONS 59

(1)

then there is a unique function Y which is a solution of (1) on all of (a, b)such that

y(xo) = Yo, y'(xo) = Yb, ... , y(n-l)(xo) = ybn-l).

The thing to note about this theorem is that it guarantees the existenceof a solution on the whole interval on which the coefficient functions arecontinuous. Moreover, the uniqueness of the solution is an automaticconsequence of the form of the equation (compare with the examples ofSection 1-3).

We saw in Chapter 1 that the uniqueness aspect of such a theorem canbe used to show that a given family of solutions contains all solutions.Theorem 1 says that there is exactly one solution Y of (1) for any initialconditions

y(xo) = Yo, y'(xo) = Yb, ... , y(n-l)(xo) = ybn-l). (3)

If we find a family of solutions of (1) such that any initial conditions (3)are satisfied by some member of the family, then the family contains allsolutions. This idea is central in our development of the theory of linearequations.

PROBLEMS

1. The equ.ation y' 1 + y2 is not linear, but the function F(x, y) = 1+ y2has all the continuity properties one could ask for. Show that nevertheless thereis no function which is a solution on the interval [0, 4]. [Hint: Solve theequation.]

2. Suppose q(x) == 0 in (1) and y is a solution of (1) on (a, b) such that forsome Xo in (a, b).

y(xo) = y'(xo) = ... = y(n-l)(xo) = O.

Show that y(x) == 0 on (a, b).3. Write out Theorem 1 specifically for the case n = 1 and prove it

(d. Section 2-4).4. Write out Theorem 1 specifically for the case n = 2.

3-2 Two theorems on linear algebraic equations. In trying to satisfyinitial conditions for solutions of a linear differential equation we are forcedto consider simultaneous linear algebraic equations. The theorems thatare needed from algebra are given here.

Suppose aij for i = 1,2, ... ,n and j = 1,2, ... ,n are given numbers,and bl , ... ; bn are given numbers. We want to know when there arenumbers CI, ... , Cn satisfying the equations

Clal1 + C2a l2 + + cnaln = bI,

Cla21 + C2a22 + + cna2n = b2,

60

The number

LINEAR EQUATIONS

all al2 alna21 a22 a2n

D=

[CHAP. 3

(2)

is called the determinant of the system (1).

THEOREM 1. If D ~ 0, there is a unique set of numbers CI, ... , ensatisfying (1).

If the numbers bl, ... , bn in (1) are all zero, then (1) is called a systemof homogeneous equations. Such a system always has at least the solutionCI = C2 = ... = Cn = O. This is called the trivial solution, and we aregenerally interested in whether there are other (nontrivial) solutions.

THEOREM 2. If b i = b2 = ... = bn = 0, then (1) has a nontrivialsolution (some Ci ~ 0) if and only if D = O.

PROBLEMS

1. Show that half of Theorem 2 is a consequence of Theorem 1.2. Consider the family of functions Clex + c2e-x • For what values of x is

there a function y in this family such that y(x) = 1, y'(x) = 2?3. Consider the family of functions Clex + c2e-x + C3 cosh x. For what values

of x is there a function y in this family such that y(x) = 1, y' (x) = 0, andy" (x) = O?

4. Solve the system3Ci + C2 + C3 = 8,CI - 2C2 + C3 = 0,

CI + C2 - C3 = 0.

5. Find a nontrivial solution of the system

Cl + 2C2 + C3 0,2Cl - C2 - C3 0,

Cl + 7C2 + 4C3 = 0.

ANSWERS

2. All values of x3. No values of x, since y"(x) == y(x) for every function in the family.4. Ci 1, C2 2, C3 = 3

5. Cl = 1, C3 = -3, C3 = 5

3-3] GENERAL THEORY OF LINEAR EQUATIONS 61

3-3 General theory of linear equations. We will consider the equation

yen) + Pn_I(X)y(n-lJ + ... + PI(X)Y' + Po(x)y = q(x),

where Po, PI, ... , Pn-l> q are defined and continuous on some interval(a, b). For simplicity, the independent variable x will be omitted from thecoefficient functions Po, ... , Pn-l, q as well as from the dummy functionY, and the equation will be written

yen) + Pn_Iy(n-l) + ... + PlY' + PoY = q.

The equation obtained from (1) by replacing q by zero,

yen) + Pn_Iy(n-l) + ... + PlY' + PoY = 0,

(1)

(2)

is called the reduced equation for (1), or the homogeneous equaMon associatedwith (1).

In this section we will not be concerned with specific solutions to specificexamples of (1). Instead we investigate the general structure of the solutions of (1) and find that there is a simple and elegant theory inherent inthe form of (1). We will show that to solve (1) it is necessary to find onlyone solution of (1) and all solutions· of the reduced equation (2). Becauseof the linear form of (2), and the fact that the right-hand member is zero,any linear combination of solutions of this equation is again a solution.Using Theorem 1 of Section 3-1, we show that the n-parameter familywhich is the set of all linear combinations of any n "essentially different"solutions of (2) is the family of all solutions.

We proceed to fill in the details.

THEOREM 1. If Yo and YI are any solutions of (1), then YI = Yo + u,where u is a solution of (2).

Proof. Define u = YI - Yo. Then we must show that u is a solutionof (2). Since u' = y~ - vb, and u" = y~' - Y{{, etc., we obtain, uponsubstituting u in the left side of (2),

(yin) - y\,n)) + Pn-l (yin-I) - y\,n-l)) + ... + PO(YI - Yo)

(yin) + Pn_Iyin- l ) + ... + POYl)

- (y\,n) + Pn_Iy\,n-l) + ... + PoYo)

= q - q = O.

THEOREM 2. If Yo is a solution of (1) and u is a solution of (2), thenYo + u is a solution of (1).

Proof. Problem 1.

62 LINEAR EQUATIONS [CHAP. 3

The following is a restatement of Theorems 1 and 2:

If Yo is any solution of (1), then the set of solutions of (1) consists of allfunctions of the form Yo + u, where u is any solution of (2). That is, tofind all solutions of (1) it 1'S necessary and sufficient to find one solution Yoof (1) and all solutions u of (2).

The following example illustrates the ideas which are developed in thesequel.

EXAMPLE 1. Consider the second order equation y" - y = 1 - x.The function yo(x) = x-I is a solution. The reduced equation isy" - y = O. By direct verification (Problem 2) one sees that clex + C2e-xis a solution of the reduced equation for any numbers Cl and C2' Todetermine that the family clex + C2e-x contains all solutions of the reducedequation, we check to see that any initial conditions y(xo) = Yo,y'(xo) = y~ are satisfied by some function in the family. For any numbersxo, Yo, y~ there must be numbers Cl and C2 such that

Clexo + C2e- xo = Yo,

clexo

- C2e- xo = y~.

The determinant of this system is

IeXo e-xo IeXo _e-xo

-2 ~ 0,

so there is a solution for Cl and C2. Theorem 1 says there is exactly onesolution for any given initial conditions, and since there is one from thefamily clex + C2e-x, this family contains all solutions of the reducedequation. The solutions of y" - y = 1 - x are therefore the functionsx-I + clex + C2e-x.

The following theorem allows us to look separately at the individualterms of the right-hand member q in finding a single solution of (1).

THEOREM 3 (Superposition principle) .. If q = ql + q2, and Yi is a solution of y(n) + Pn_ly(n-ll + ... + PoY = qi, (i = 1,2), then Yl + Y2is a solution of (1).

Proof. If Yl + Y2 is substituted in the left side of (1), the terms containing Yl and its derivatives can be separated from those containing Y2and its derivatives. By hypothesis, the terms containing Yl add up to ql,and those containing Y2 add up to q2.

Now we turn to the structure of the solutions of the reduced equation (2).


THEOREM 4. If Yl, ... , Yk are solutions of (2), then any linear combination ClYI + ... + CkYk (Cl> ... , Ck constants) is also a solution of (2).

Proof. The theorem follows by induction once it is proved for any twofunctions and any two constants. So assume YI and Y2 are solutions of (2)and let u = CIYl + C2Y2' Then

U" = CIY'! + C2Y~, etc.

Substituting u in (2), we get

u(n) + Pn_lU(n-l) + ... + PoU

(clyin) + c2y~n» + Pn_I(CIyin- l ) + c2y~n-I» + ...+ PO(ClYI + C2Y2)

= cI[yin) + Pn_Iyin- l ) + ... PoYtl

+ C2[y~n) + Pn_Iy~n-l) + ... + POY2]

= cIO + C20 = O.

EXAMPLE 2. The functions eX, e-x, sinh x are solutions of Y'" - Y' = O.

By Theorem 4, any linear combination clex + c2e-x + Ca sinh x is also asolution. However, anyone of the functions eX, e-x

, sinh x can be writtenas a linear combination of the other two (e.g., sinh x = !eX

- !e-X),

and the family of linear combinations is in reality a two-parameter family,

where C I = CI + !ca and C2 = C2 - !ca. A set of initial conditionsy(xo) = Yo, Y' (xo) = Yb, Y" (xo) = Yb' for this family would consist ofthree linear equations in the two unknowns CI and C2 :

y'(xo) = Clexo- C2e-xo = y~,

y"(xo) = CIexo + C2e-xo = y'r).

Such a system will not always have a solution, and the family does notcontain all solutions.

Now consider in general the problem of satisfying any initial conditionswith some function from a given family. Suppose YlI ... , Yn are solutionsof the homogeneous equation (2), so that each function in the family

Y = CIYl + ... + CnYn (3)


is also a solution. Consider any set of initial conditions

Y(Xo) = Yo, (4)

If one of the functions (3) is to satisfy (4), we must have numbers Cll ... , Cn

satisfying

= Yo,

(5)CIy'1 (xo) + + cnY~(XO) = Yb,

cIYln-I)(xo) + + cny~n-l'(xo) = ybn- lJ •

According to Theorem 1, Section 3-2, the system (5) is satisfied for somenumbers Cll ... ,Cn provided the determinant of the coefficients is notzero. For any set of (n - 1) times differentiable functions YI, ... , Yn, thedeterminant

YI(X)

W(YI(X), ... , Yn(x)) = y)(x) y~(x)

Yln-l)(x) y~n-l)(x)

Yn(x)

y~(x)

y~n-l)(x)

is called the Wronskian of the functions. Restating the remarks above,we see that the system (5) will have a solution for CI, ... , Cn providedW(YI (xo), ... , Yn(Xo)) ,t. O. If the family (3) is to satisfy every set of initialconditions (in particular, every xo), we must have W(YI (x), . .. ,Yn(X)) ,t. 0for all x. The situation is actually somewhat simpler than indicated, asthe following definition and theorems show.

DEFINITION 1. Functions UI, ... , Uk are linearly dependent on aninterval I if there are numbers Cll ... ,Ck, not all zero, such thatCIUI(X) + ... + CkUk(X) == 0 on I. The functions UI, ... , Uk arelinearly independent on I if they are not linearly dependent; i.e., ifCIUI(X) + ... + CkUk(X) == 0 implies CI = C2 = ... = Ck = O.

EXAMPLE 3. Linearly dependent functions.

(A) eX, e-x , cosh Xi !eX+ !e-X - cosh x == 0

(B) 0, x, eX; 14·0 + 0 . x + 0 . eX == 0

(C) x + 1, 2x - 3, 5; 2(x + 1) - (2x - 3) - 5 == 0

(D) 2, sin 2 x, cos2 x; 1 ·2 - 2· sin 2 x - 2· cos2 X == 0

Linearly independent functions.

(E) 1, x, x 2, x 3

(F) eX, e2"', e3x

(G) eX, xex , sin x

(H) 1, cos x


THEOREM 5. If Ul, ... , Uk are any (k - 1) times differentiable functionswhich are linearly dependent on I, then W(Ul(X), ... ,Uk(X)) = °on I.

Proof. Let CIUl(X) + ... + CkUk(X) = °on (a, b) with not all Ci = 0.Differentiating (k - 1) times, we obtain the relations

CIUl (x) + + CkUk(X) = 0,

CIUi(X) + + ckuHx) = 0,

cluik-1\X) + + ckuick-I)(X) = 0.

(6)

For any fixed x, this is a system of homogeneous algebraic equations inCI, ... ,Ck. By assumption, this system has a nontrivial solution for each x.By Theorem 2, Section 3-2, the determinant W(UI(X), ... ,Uk(X)) = °for all x.

COROLLARY. If the Wronskian of any set of functions is nonzero forany Xo., then the functions are linearly independent on any interval containing xo.

The converse of Theorem 5 is false for arbitrary functions Ul, ... , Uk(Problem 7). However, if we have n functions which are solutions of thenth order homogeneous equation (2), a statement even stronger than theconverse of Theorem 5 is true.

THEOREM 6. If Yl, ... , Yn are n solutions of the nth order homogeneousequation (2), and W(Yl(XO),"" Yn(Xo)) = ° for some Xo in (a, b),then Yl, ... ,Yn are linearly dependent on (a, b), and hence W(YI(X), ... ,Yn(x)) = °on (a, b).

Proof. Recall (Problem 2, Section 3-1,) that the only solution of (2)satisfying y(xo) = y'(xo) = ... = y(n-lJ(xo) = °is the function identically zero. Now suppose W(YI(XO), ... ,Yn(Xo)) = 0. Then the system

CIYl (xo)

CIyi(XO)

+ + cnYn(XO)

+ + cnY~(XO)

= 0,

= 0,(7)

has a nontrivial solution, Cl, ... ,Cn not all zero. For any Cl, •.. , Cn

which are a nontrivial solution of (7), let Y = CIYI + ... + CnYn' ThenY is a solution of (2), and by (7), we have y(xo) = y'(xo) = =y(n-lJ(xo) = 0. Therefore Y is identically zero, which says that Yl, , Yn


are linearly dependent on (a, b). By Theorem 5 the Wronskian ofYll ... , Yn is identi?ally zero.

COROLLARY. The Wronskian of n solutions of (2) is identically zero,or is never zero.

THEOREM 7. The nth order equation (2) has n solutions which are linearlyindependent on (a, b). If Yll ... , Yn is any set of linearly independentsolutions of (2), then the family

contains all solutions.

ClYl + ... + CnYn (8)

Proof. The fact that (8) contains all solutions is immediate from theCorollary to Theorem 6. The fact that the Wronskian is never zero forindependent solutions allows us to satisfy any set of initial conditions asin (5). The uniqueness statement of Theorem 1, Section 3-1, shows thatthe solution for any set of initial conditions is in the family (8), and thus (8)contains all solutions. To see that there are n independent solutions, consider the n solutions Yl, ... , Yn corresponding to the n sets of initial conditions (i), ... , (n):

(i) y(xo) = 1, y'(xo) = y"(xo) = ... = y(n-l\xo) = 0,

(ii) y(xo) = 0, y'(xo) = 1, y"(xo) = ... = y(n-ll(xo) = 0,

The Wronskian of Yll ... , Yn at Xo is

1 0 0 0

010 o1,

o 0 0 1

so these solutions are independent.We will refer to (8) as the general solution of the homogeneous equation

(2). Any specific function Yo which is a solution of (1) will be called aparticular solution of (1). The expression

Yo + C1Y + ... + cnYn, (9)

which is the set of solutions of (1), will be called the general solution of (1).

3-3] GENERAL THEORY OF LINEAR EQUATIONS

PROBLEMS

67

1. Prove Theorem 2.2. Verify (see Example 1) that cle" + C2e-x is a solution of y" - y = 0 for

all numbers CI, C2.

3. Show, as in Example 1, that every function in the family clex + C2xeX isa solution of y" - 2y' + y = 0, and that the family contains all solutions.

4. The functions 2 + x, x - eX, and x + 1 + eX are solutions of a certainnonhomogeneous second order linear equation.

(a) Find (Theorem 1) two solutions of the reduced equation, neither ofwhich is a constant multiple of the other.

(b) Write a two-parameter family of solutions of the reduced equation(Theorem 4) and show that it contains all solutions of the reducedequation (cL Example 1).

(c) Find all solutions of the given nonhomogeneous equation (Theorems1 and 2).

5. (See Example 2.) Find by inspection a third solution U of y'" - y' = 0which is not a linear combination of eX and e-x. Show that the family CIU +c2ex + cae-x contains all solutions.

6. (a) Show that Ul, .•. , Uk are linearly dependent if and orily if some Uj

can be written as a linear combination of the remaining functions.(b) Find three functions UI, U2, ua which are linearly dependent and such

that UI cannot be expre:;;sed as a linear combination of U2 and Ua.

(c) Show that any set of functions containing the function identically zerois linearly dependent.

7. Show that x3 and Ixl 3 are not linearly dependent on [-1,1], but thatW(x3 , Ix1 3) == O. This shows that the converse of Theorem 5 is false.

8. Are x 3 and Ixl 3 solutions on [-1, 1] of (a) any second order homogeneouslinear equation? (See problem 7.) (b) any third order homogeneous linearequation? (What is the third derivative of Ixl 3 at O?)

9. Which of these sets of functions are linearly dependent on the whole line?

(a) x - 3, 6 - 2x, eX (b) x + 1, 2x - 3, 3x + 4(c) cos2 x, cos 2x, sin2 x (d) 1, ex, xeX

10. (a) Compute the Wronskians of the functions in (E), (F), (G), and (H) ofExample 3.

(b) Show (Theorem 6) that the functions in (H) are not solutions on[-1,1] of any second order homogeneous linear equation.

(c) The functions of part (G) of Example 3 are solutions of y"" - 2y'" +2y" - 2y' + y = O. The Wronskian vanishes (at 7r/2) but is notidentically zero. Why doesn't this contradict Theorem 6?

11. Suppose YI, ... , Yn are linearly independent solutions of (2), and Yo is asolution of (1). Show directly that any initial conditions are satisfied by somefunction in the family yo + ClYI + ... + CnYn.

12. Show that any n + 1 solutions of an nth order homogeneous linear equation are linearly dependent.

68 LINEAR EQUATIONS

ANSWERS

[CHAP. 3

4. (a) 2 + e", 1 + 2e"; the condition of the problem rules out the choice ofthe zero function.

(b) CI + C2e"; any initial conditions can be satisfied with a function fromthis family.

(c) x + C1 + c2e"5. u(x) = 16. U2 andu3 must be linearly dependent8. (a) No, (b) No9. (a), (b), and (c)

10. (E) 12, (F) 2e6", (G) -2 cos xe2", (ll) -sin x

3-4 Second order equations with constant coefficients; One of thesimplest and most useful cases of the linear equation is the second orderequation

Y" + PlY' + P2Y = q(x), (1)

where PI and P2 are constants. We will examine this equation in somedetail and apply our results in Section 3-5 to some examples from mechanics, electricity, etc.

First, let us look at the reduced equation

y" + PlY' + P2Y = o. (2)

The problem is to find two linearly independent solutions. Since twofunctions are linearly dependent only if one is a constant multiple of theother (Problem 1), the check for independence can be made by inspection.

Substituting Y = eTX in the left side of (2), we get

Hence eTX is a solution of (2) if r is a root of the algebraic equation

r2 + Plr + P2 = O.

(3)

(4)

The equation (4) is called the auxiliary equation for (1) or (2). If (4) has.two real roots, rl and r2, then eT1X and eT2X are the required two solutionsof (2). If (4) has only one real root, ro, then

r2 + Plr + P2 = (r - ro)2 = r2 - 2ror' + r~.

That is, PI = -2ro and P2 = r~. In this case the solutions of (2) areeToX and xeToX (Problem 2). If (4) has no real roots, the roots are conjugate

3-4] SECOND ORDER EQUATIONS, CONSTANT COEFFICIENTS 69

complex numbers, say a + ib, a - ib, and

r2 + Plr + P2 = [r - (a + ib)][r - (a - ib))

= r2 - 2ar + a2 + b2.

In other words,*

(5)

PI = -2a and (6)

One can verify by substitution (Problem 3) that the solutions of (2) mthis case are eax cos bx and eax sin bx.

To summarize, the family of solutions of (2) is

CleT1X + C2eT2X,cleTox + C2xeT0X,

cleax cos bx + C2eax sin bx,

if r I and r2 are distinct real roots of (4),

if 1'0 is the only real root of (4),

if a ± ib are the roots of (4).

EXAMPLE 1. (A) y" + y' - 2y = O.

The auxiliary equation is r2 + r - 2 = 0, or (r + 2)(r - 1) = O. Theroots are 1 and -2, so the solutions of (A) are clex + C2e-2x.

(B) y" + 4y' + 4y = O.

The auxiliary equation is r2 + 4r + 4 = 0, or (r + 2) 2 = O. The singleroot is -2, so the solutions of (B) are y = cle-2x + C2xe-2x.

(C) y" - 2y' + 5y = O.

The auxiliary equation is r2 - 2r + 5 = O. Since the discriminant is(_2)2 - 4(1)(5) = -16 < 0, the roots are complex numbers a ± ib.Comparing with (6), we see that a = -!(-2) = 1, and b = vl5 - 12 =2.The solutions of (C) are y = clex cos 2x + C2ex sin 2x.

Having found all the solutions of (2), we are still faced with the problemof finding one solution of (1). Later we will give the so-called variation ofparameters method for finding a solution of any linear equation when allthe solutions of the reduced equation are known. For now we consider thesimpler method of undetermined coefficients, which works whenever theright member q of (1) has the form

P(x)eax cos bx + Q(x)eax sin bx, (7)

* Note that formulas (6) give an easy way to find a and b when (4) has thecomplex roots a ± ib; a = -!PI, and b = VP2 - a2 •


with P and Q polynomials. Some examples of functions of the form (7)are:

x 3- 2x + 1

(2 - x)eX

3 cos 2x

x 2 sin x

3 cos x - x sin x

2xeXcos 2x

(a = b = 0, P(x) = x 3- 2x + 1),

(a = 1, b = 0, P(x) = 2 - x),

(a = 0, b = 2, P(x) = 3, Q(x) = 0),

(a = 0, b = 1, P(x) = 0, Q(x) = x 2),

(a = 0, b = 1, P(x) = 3, Q(x) = x),

(a = 1, b = 2, P(x) = 2x, Q(x) = 0).

The facts behind the method of undetermined coefficients are stated inthe following theorem, which is proved later as a simple consequence ofthe theory of operators.

THEOREM 1 (Method of undetermined coefficients). Let Pn, Qn, P~, Q~

be polynomials of degree n or less.If q(x) = Pn(x)eaX, then (1) has a solution of the form y = xkP~(x)eaX,

where k is 0, 1, or 2; k = 0 if eax is not a solution of the reduced equation(2); k = 1 if eax is a solution of (2) and xeax is not; k = 2 if both eax

and xeax are solutions of (2).If q(x) = Pn(x)eaX cos bx, or q(x) = Qn(x)eaX sin bx, or q(x) =

Pn(x)eaX cos bx + Qn(x)eaX sin bx, with b ~ 0, then (1) has a solutionof the form y = xk(P~(x)eaX cos bx + Q~(x)eaX sin bx) where k is 0 or 1;k. 0 if eax cos bx is not a solution of (2), and k = 1 if eax cos bx is asolution of (2).

The "method" consists simply of substituting in the equation a functionof the appropriate form with arbitrary polynomials P~(x) = A o +Alx + ... + Anxn, Q~(x) = B o + Blx + ... + Bnxn, and seeing whatthe coefficients A o, AI, ... ,Bo, B I , ... must be for the function to bea solution. Note that if q contains either a sine or a cosine term, thetrial function must contain both, and the polynomial coefficients P~,

Q~must both be of the degree which is the maximum of the degrees ofPn and Qn. The conditions on the factor xk can be remembered as follows.A trial solution of the same form as q is multiplied by x or x 2

, if necessary,so that none of the terms of the resulting function are solutions of the reduced equation. For example, if q(x) = xex, the trial solution of the sameform is (A + Bx)ex. If eX is a solution of the reduced equation, there is nopoint in including the term Aex in the trial function, and the appropriatefunction is (Ax + Bx2 )ex. If both eX and xeXare solutions of the reducedequation, the trial function should be

3-4) SECOND ORDER EQUATIONS: CONSTANT COEFFICIENTS 71

EXAMPLE 2. (i) y" - y' - 2y = 1 - x2,

(ii) y" - y' = 1 - x 2•

The solutions of the reduced form of (i) are cle-x + C2e2x. There is asolution of (i) of the form

y = (A + Bx + Cx2)eOX = A + Bx + Cx2.

Substitution gives

2C - (B + 2Cx) - 2(A + Bx + Cx2)or

(2C - B - 2A) + (-2B - 2C)x - 2Cx2.

We must have -2C = -1, -2B - 2C = 0, and 2C - B - 2A = 1.This gives C = i, B = -i, and A = l The solutions of (i) are

y = i- - !x + ix2 + cle-x + C2e2x.

The solutions of the reduced form of (ii) are Cl + C2ex. Since constantsare solutions of the reduced equation, the trial solution for (ii) is y =Ax + Bx2 + Cx3

. The coefficients A, B, C are evaluated as above.

EXAMPLE 3. (i) y" - y' - 2y = eX,

(ii) y" - y' - 2y = (1 + x2)eX,

(iii) y" - 3y' + 2y = 2ex,

(iv) y" - 2y' + y = (2 + x)ex.

In (i) and (ii), eX is not a solution of the reduced equation, and the trialsolutions are respectively y = Aex, and y = (A + Bx + Cx2)ex. In (iii)eX is a solution of the reduced equation, and the trial solution is y = Axex.In (iv), eX and xeXare solutions of the reduced equation, so the trial solution is y = x 2(A + Bx)ex. For example, in (iii), substitution of y = Axex

gives A(2 + x)eX - 3A(1 + x)eX+ 2Axex = -Aex. Therefore Axex isa solution if A = -2, and the set of all solutions is (Cl - 2x)eX+C2e2x.

EXAMPLE 4. (i) y" - y' - 2y = cos x,

(ii) y" + y = cos x,

(iii) y" + y = x cos x.

Here cos x and sin x are not solutions of the reduced equation in (i),but are in (ii) and (iii). The trial solutions for the three equations are

(i) y = A cos x + B sin x,

(ii) y = Ax cos x + Bx sin x,

(iii) y = (Ax + Bx2) cos X + (Cx + Dx2) sin x.


EXAMPLE 5. (i) y" + y = eX sin 2x,

(ii) y" - 2y' + 5y = 3xex cos 2x,

(iii) y" - 2y' + 5y = xex cos 2x - eX sin 2x.

Here eX cos 2x and eX sin 2x are not solutions of the reduced equationfor (i), but are for (ii) and (iii). The trial solutions are

(i) y = Aex cos 2x + Bex sin 2x,

(ii) y = (Ax + Bx2 )eXcos 2x + (ex + Dx 2)eXsin 2x,

(iii) same as (ii).

The superposition principle (Theorem 3, Section 3-3) can be used toextend the method of undetermined coefficients to any equation in whichthe right member is a sum of functions of the form (7).

EXAMPLE 6. y" - y' - 2y = 1 - x 2 + 2eX - cos x.

We consider separately the equations y" - y' - 2y = 1 - x 2 [Example2(i»), y" - y' - 2y = 2eX[Example 3(i»), and y" - y' - 2y = -cos x[Example 4(i)]. Solutions of these three equations are respectively! - !x + !x2

, -ex, and 130 cos x + lo sin x. The solutions of the given

equation are therefore

PROBLEMS

1. Show that two functions are linearly dependent if and only if one is aconstant multiple of the other.

2. Show that if (4) has only one real toot TO, then erox and xerox are solutionsof (2).

3. Show that if (4) has the roots a + ib and a - ib, so that (2) has the formy" - 2ay' + (a2 + b2)y = 0, then eax cos bx and eax sin bx are solutions of (2).


(a) y" + y' + y = 0

(b) y" - 3y' + 2y = 0

(c) y" + y' - 6y = 0

Find all solutions for the following and specify which solution satisfies thegiven initial conditions.

5. (a) y" = 0, y(2) = 2, y' (2) = 3

(b) y" - 2y' + y = 0, yeO) = 1, y'(O) 2

(c) y" + 4y = 0, yeO) = 1, y'(O) = -2

(d) y" - 2y' = 0, yeO) = l,y'(O) = 4

3-4] SECOND ORDER EQUATIONS: CONSTANT COEFFICIENTS 73

6. y" + 3y' = 6x, yeO) = 1, y'(O) = i7. y" + y = e"', yeO) = 1, y'(O) = 0

8. y" - 2y' + y = -25 sin 2x, yeO) = -4, y'(O) = 8

9. y" - 2y' + y = e"', y(l) = ie, y'(!) = -ie

Find all solutions for Problems 10 through 14.

10. y" - 3y' + 2y = x2 + 1

12. y" + y' - 2y = e'"

14. y" - 2y' = 4x + e3",

11. y" + y' + y = x + 2 + 3e'"

13. y" - 2y' + 2y = e'" cos x

Write the form of a single solution for each of the following, but leave theeoefficients undetermined.

15. y" - 4y = (x3 - 2x)e2", + x cos x

16. y" + 2y' - 3y = xe'" + e-3", sin x + e'"

17. y" + 4y' + 8y ;" xe-2"'(3 + sin 2x)

18. (a) Prove that if q(x) = ao + alX + + anxn, and P2 ~ 0, then (1)has a solution y = Ao + AIX + + Anxn. [Hint: Start with thecoefficient of xn after substitution and work backwards.]

(b) Consider the cases P2 = 0, PI ~ 0, and PI = P2 = O.(c) Show that if q(x) = (ao + alX + ... + anxn)eQ

"', the substitutiony(x) = ea",z(x) changes (1) to a linear equation in z with a polynomialright member. Use this and parts (a) and (b) to prove the first halfof Theorem 1.

ANSWERS

4. (a) Y-(112)", V3 + -(112)",. V3

= Cle cos "2 x C2e sm"2 x

(b) y = cle'" + c2i'" .(c) y = cIl'" + c2e-3

",

5. (a) y = CI + C2X, Y = -4 + 3x(b) y = (CI + c2x)e"', y = (1 + x)e'"(c) y = CI cos 2x + C2 sin 2x, y = cos 2x - sin 2x(d) y = CI + C2e2"" y = -1 + 2e2",

6. y CI + c2e-3", ~ ~x + x2, Y = 2 - e-3", - ~x + x2

7. y = Ci cos x + C2 sin x + !e"', y = !(e'" + cos x - sin x)8. y = (Ci + c2x)eX - 4 cos 2x + 3 sin 2x,

y -4 cos 2x + 3 sin 2x + 2xe'"9. y (CI + C2X + !x2)e"', y = (1 + !x2)e'"

10. y = :£ + ix + !x2+ cle'" + c2e2",

"'+ -(112)",( V3 . V3 )11. Y = 1 + x + e e CI cos "2 x + C2 sm"2 x


12. Y (Cl + !x)ex + c2e-2x

13. y !xeXsin x+ eX(cl cos x + C2 sin x)14. y -x - x2 + !e3x + Cl + c2e2x

15. y = x(A + Bx + Cx2 + Dx3)e2X + (E + Fx) cos x + (G + Hx) sin x16. y = x(A + Bx)ex + Ce-3x sin x + De-3x cos x17. y = x(A + Bx)e-2x sin 2x + x(C + Dx)e-2X cos 2x + (E + Fx)e-2x

3-5 Applications. Many of the common differential equations whicharise in applications are of the type

Y" + PlY' + P2Y = q(x) (PI, P2 constants), (1)

which was treated in the preceding section. In this section we will examinethe differential equations of some simple physical systems and try tointerpret the solutions in physical terms.

As a first example, consider a block of mass m attached to a spring andsliding on a horizontal table (Fig. 3-1). Let the spring constant be b2 lb/ftso that the spring exerts a force of -b2s lb when the spring is stretched(s > 0) or compressed (s < 0) s ft. Assume that the assorted frictionaleffects exert a force proportional to the speed s' = ds/dt, and of courseopposing the motion. This so-called damping force can be written -2as',with a ~ O. Since mass times acceleration (s" = d2s/dt2

) equals thetotal force, we have the equation ms" = -2as' - b2s, or

If2a' b2

s" + - s' + - s = O. (2)m m

(3)

/.C.. 2\ .

s" + 2a s' + ~·s = 1.. F(t).m m m

If in addition some external force F(t) acts on the mass, the equationbecomes

As an example of (3), let the mass hang from the spring (Fig. 3-2), sothere is a constant force F(t) = mg exerted by gravity.

The equation (3) also appears in the study of electric circuits. The current i in a series circuit containing a resistance R, inductance L, and

Motion-

1-Natural-I-s--llength . I Forces:

F(t)_-b2s--2as'

FIGURE 3-1

3-5) APPLICATIONS 75

capacitance C is determined by the equation

(4)

(5)E' = F,1

C = b2 'L= m,R = 2a,

where E(t) is the electromotive force applied at time t. If we have theequalities

(6)mm

then Eqs. (3) and (4) are identical except for the dummy function used.This formal similarity between electric circuits and vibrating mechanicalsystems makes it possible to study the behavior of a mechanical systemby setting up the corresponding circuit {by (5)] and measuring the currentdirectly.

Now let us examine the solutions of (2), which will be added to anyparticular solution of (3), or with the changes (5), to any solution of (4).Physically, the solutions of (2) represent the motion which results if thesystem of Fig. 3-1 is set into motion and then released with no furtherforce applied. These solutions are called transients, since the motion theyrepresent dies out as t increases. The roots of the auxiliary equation for (2)are r-'

-a + Va2 - b2m -=-a - Va 2 - b2mand

The nature of the solutions will depend on the sign of a2- b2m; that is,

on the relative magnitudes of the damping force and the restoring forcetimes the mass. Let us consider separately the cases a2 > b2m,a2 = b2m, and a2 < b2m.

Case I (a2 > b2m; overdamping). The roots (6) are distinct realnumbers, and the solutions of (2) are the functions

(-a + va2

- b2m) (-a - Va 2- b2m)

8 = Cl exp t + C2 exp . t.m m (7)

Motion t Forces: mg

~

t

FIGURE 3-2


Since a > y'a2- b2m, both terms in (7) are of the form ce-pt

, p > 0,and tend monotonically to zero as t ~ 00. The functions (7) change signonce or not at all (Problem 1), depending on the initial conditions.

Case II (a2 = b2m; critical damping). The auxiliary equation has onereal root, -aim, and the solutions of (2) are

(8)

The exponential is always positive, so each function (8) changes sign atmost once. In both Cases I and II the damping force predominates, andthere is no oscillation. The mass of Fig. 3-1 crosses over the equilibriumpoint at most once, then returns toward the equilibrium position. Thiskind of behavior is illustrated by the wheels of an automobile acting underthe influence of the springs and shock absorbers.

. Case III (a2 < b2m; underdamping). The roots (6) of the auxiliaryequation are the complex numbers -aim ± iw, where

1w = - Y mb 2 - a2 •

m(9)

(10)

(11)C2COS a = -;:.:::;::::==:::::

YC2 + c21 2

The solutions of (2) are

s = e-(a/m)t[Cl cos wt + C2 sin wt].

If we define new constants A and a by

. A = Yc2 + c2 sin a = Cl1 2' YC2 + c2

1 2

then (10) can be written

s = Yc2 + c2 e-(a/m) t { Cl cos wt + C2 sin wt}1 2 Y c2 + c2 Y c2 + c2

1 2 1 2 (12)= Ae-(a/m)t sin (wt + a).

The solutions are damped sine curves (Fig. 3-3). The motion is an oscillation about the equilibrium point with the amplitude of the vibrationsdecreasing to zero as t increases. The period of the vibration, or time required for a complete cycle, is 2'rr/w sec. The frequency is w/27r cps.

In case the damping effects are negligible, a = 0, we have what is calledsimple harmonic motion. The equation (2) in this case is

b2

8"+-S=0m '

(13)

3-5]

s

APPLICATIONS 77

FIGURE 3-3

and the solutions are pure sine curves

bt . bt8 = CI cos vm + Cz sm vm'

or

(14)

Now let us consider the situation in which the vibrations are forced,as in (3) or (4). The solutions we have found for (2) describe transienteffects; the motion they represent is superimposed on the motion (a particular solution) which corresponds to the forcing function. The type offorcing function which can realistically be considered in (3) or (4) is ofcourse limited; an unbounded function, for example, would not be reasonable. We will consider a forcing function of the form F(t) = E sin (wot)which is the case, for iristance, when an alternating electromotive force isapplied to the circuit (4). The equation we consider is therefore

2a bZ E.8" + - 8' + - 8 = - sm (wot).

m m m

By Theorem 1, Section 3-4, there is a solution of (15) of the form

8 = Al cos (wot) + A z sin (wot),

(15)


or equivalently, of the form

8 = A sin (wot - a).

Substitution of (16) in the left side of (15) gives

{b2 - mw~ . 2awo }A m sm (wot - a) + ----:m:- cos (wot - a) "

Now let

1= - v(b2 - mw2) 2 + 4a2w2mOo'

and write (17) in the form

{b2 - mw~ . 2awo )}

AK mK sm (wot - a) + mK cos (wot - a "

The quantity K is chosen so that

(b2

- mw~)2 + (2awo)2 = 1mK mK '

and hence there is a number a such that

(16)

(17)

(18)

b2- mw~

cos a = mK

For this value of a, (18) becomes

and. 2awo

sma= mK" (19)

AK sin (w.ot). (20)

Therefore (16) is a solution of (15) provided a is determined by (19) andAK = Elm; that is, if

(21)

For a forcing function F(t) = E sin (wot), the solutions will all approximate,as the transients die out, a sine curve with the same frequency wo/27ras the forcing function. The maximum displacement, as one would expect,will not occur at the same time as the maximum force, but will lag by aninterval of alwo seconds.

3-5) APPLICATIONS 79

If there is little damping (a is small) and the frequency wo/27r of thedriving force approaches the natural damped frequency w/27r [Formula (9)]of the system, then the amplitude A can be very large. (See Problem 3.)This is the phenomenon called resonance.

EXAMPLE 1. (See Fig. 3-2.) If a block weighing 32lb is suspendedfrom a spring, the spring stretches 16 in. (t ft). Suppose the damping forcein pounds equals twice the numerical value of the speed in feet per second.Find the position of the block at time t if the block is released at t = °withzero velocity and with the spring at its natural length.

The equation is (3) with F(t) = 32, 2a = 2, and m = 1 (takingg = 32 ft/sec2). The spring constant b2 is given by tb 2 = 32. Henceb2 = 24, and the equation is

s" + 2s' + 24s = 32,

with the initial conditions

s(o) = 0, s'(O) = 0.

The roots of the auxiliary equation are

-l+V23i and -1 - V23i,

and the solutions of the reduced equation are

A particular solution of the equation is clearly the constant s = t. Whent = 0, s = t + Cl = 0, so Cl = -l The velocity at time t is given by

s' = _~[_e-t cos (V23 t) - V23 e-t sin (V23 t)]

+c2[-e- t sin (V23 t) + V23 e- t cos (V23 t)].

When t = 0, s' =. -!(-1) + c2V23 = 0, so C2 = -4/(3V23). Theposition s at time t is

s = ~ - ~e-t [cos (V23 t) + _~ sin (V23 t)].v23

EXAMPLE 2. The forces on a pendulum of mass m and length L areshown in Fig. 3-4. The force mg of gravity can be resolved into a component along the length of the pendulum, and a component mg sin e in thedirection of the motion. We measure distance s along the arc of the

80 LINEAR EQUATIONS

FIGURE 3-4

F2 -

I-S1~a-~a~

FIGURE 3-5

[CHAP. 3

pendulum, so s = LO, and s" = LO". If we assume that there IS nofriction, the equation is

ms" = -mg sin 0, or 0" + i sin 0 = o.

For small values of 0, sin 0 can usefully be approximated by 0, and we canassume the pendulum satisfies the equation

0" + f 0 = O.

This is the simple harmonic motion of (13), and the motion is periodicwith constant frequency (27r.../L7(;)-1. To study motion other than smalloscillations about 0 = 0, the exact equation must be used.

EXAMPLE 3. A bar of weight W lies across two counter-rotating cylindersas in Figure 3-5. If the coefficient of friction between the bar and the

3-5] APPLICATIONS 81

rollers is JJ., then the left cylinder exerts a horizontal force of JJ.F I, and theright cylinder exerts a horizontal force of -JJ.Fz, where F l and Fz are thevertical forces on the rollers due to the weight of the bar. Since F 1 +Fz = W, and (a + s)F l = (a - s)Fz, the equation describing the position of the bar's center of gravity is (Problem 5)

Sll + gJJ. S = 0,a

and the motion is simple harmonic.

PROBLEMS

1. Show that the functions (7) change sign once for positive t if q/Cz < -1and do not change sign if c11Cz ~ -1. [Hint: consider the function 8 = q e-pi+cze-ql, where 0 < P <q, and write s = cle-PI[I + (cz!q)e(p-q)t].]

2. Suppose the mass of Fig. 3-1 satisfies equation (2) and aZ = bZm. Express8 in terms of the initial position So and initial velocity s~. Suppose So > 0:For what initial velocities 80 will the mass fail to cross the equilibrium point?

3. Find the amplitude A [formula (11)] of the solution of (15) when thefrequency wo of the driving force equals the natural damped frequency [formula(9)] of the system.

4. Write the solution of the equation of Example 1 in the form 8 = !- +Ae-i sin (v23 t + ex); that is, find A and ex so this formula satisfies the initialconditions 8(0) = 0 = 8' (0).

5. Complete the derivation of the equation of Example 3. What is the maximum velocity the center of gravity of the bar can have midway between therollers without the bar falling off?

6. What length pendulum (Example 2) will swing from one side to the othereach second? (Take g = 32 ft/secz.)

7. A cylinder of mass m has a cross section area of 2 ftz. The cylinder floatswith its axis vertical in water of density p Ib/ft. Show that if the cylinder isdisturbed from equilibrium, it bobs up and down in simple harmonic motion.Find the frequency of the motion. (The bouyant force is equal to the weight ofwater displaced.)

8. A 16-ft chain weighing 21b/ft (total mass equals 1) rests on a table withpart of the chain hanging over the edge of the table. The coefficient of frictionbetween the chain and table is k, so that friction exerts a retarding force ofk· (16 - x) . 21b when x feet of chain are over the edge of the table. Supposethe chain is released with 6 ft hanging over the edge. How long does it take thechain to slide off?

9. A block weighing 641b hangs from a spring as in Fig. 3-2, and stretchesthe spring 6 in. The block is also attached to a shock absorber which exerts aforce of 4v Ib, where v is the speed in feet per second. If the block receives animpact giving it an initial velocity of 20 ft/sec downward at the equilibriumposition, find the position at time t. (Measure 8 downward from the naturallength of the spring, and use g = 32 ft/secz.)

82 LINEAR EQUATIONS

ANSWERS

[CHAP. 3

3. A

2. S = [SO + (Sb + a:) tJ e-<almlt;

Em

ava2 + 4m2w2o

4. A - ~ ~ == -1.36, IX ==

5. Sb = va;;;;.7.! IP cps

7l" '\j~

9. S = ~ +~ e-tsin (V63 t)

asom

1.37

6. 3.24 ft

8. ~ cosh-1 6 sec

CHAPTER 4

SPECIAL METHODS FOR LINEAR EQUATIONS

4-1 Complex-valued solutions. The treatment of homogeneous lineardifferential equations with constant coefficients is much simplified if weconsider complex solutions. We first review some of the basic facts aboutcomplex functions.

A complex-valued function f of a real variable is a rule which assigns aunique complex number to each real number x in the domain of fi thusfor each x, f(x) = u(x) + iv(x), where u(x) and vex) are real numbers. Acomplex-valued function therefore consists of a pair of real functions, uand v, which are called respectively the real part of f and the imaginarypart of f. Limits are defined for complex functions in the same way as forreal functions, only replacing the notion of distance between real numbersby the distance between complex numbers. If z = u + iv and w = r + is,with u, v, r, s real, then the distance between z and w is

Iz - wi = v(u - r)2 + (v - S)2.

DEFINITION 1. If f = u + iv is a complex function, then

lim f(x) = Uo + ivox---+xo

(1)

if and only if for each positive number E, there is a positive number 0such that If(x) - (uo + ivo)! = lu(x) + iiJ(x) - (uo + ivo)/ < E whenever 0 < Ix - xol < o.

From (1) it is clear that lu(x) + iv(x) - (uo + ivo) I is small if andonly if both lu(x) - uol and Iv(x) - vol are small. With this in mind,the following theorem follows readily from the definition above.

THEOREM 1. Iff = u + iv, then lim:z;-+:z;of(x) = Uo + ivo if and only iflim:z;-+:z;o u(x) = Uo and lim:z;-+:z;o vex) = Vo.

Proof. Problem 1.The derivative of a complex-valued function of a real variable is defined

in the same way as the derivative of a real function, and we use the samenotation.

DEFINITION 2. If f = u + iv, then we define

Df(x) = f'(x) = lim f(x + h) - f(x) .h->O h

83

84 SPECIAL METHODS FOR LINEAR EQUATIONS [CHAP. 4

The difference quotient for f can be written

f(x ~ h) -- f(x)h

u(x + h) + iv(x + h) -- u(x) -- iv(x)h

= u(x + h) -- u(x) + i vex + h) -- vex) .h h

(2)

,Applying Theorem 1 to (2), we see that both expressions

u(x + h) -- u(x)h

and vex + h) -- vex)h

must approach limits if f is differentiable. Therefore f is differentiable ifand only if both u and v are, and we have the formula

Df(x) = f'(x) = u'(x) + iv'(x). (3)

The familiar theorems for derivatives of real functions go over unchangedto complex functions (Problem 2). The sum, product and quotient rules,and the chain rule, are exactly the same for real and complex functions.For example,

D(j(x)g(x» = f(x)g'(x) + f'(x)g(x) ,

andDf(g(x» = f'(g(x»g'(x),

whether f and g are real or complex.

EXAMPLE 1.

(i) D[x2 + x + i(x3 + cos x)] = 2x + 1 + i(3x2-- sin x)

(ii) D[eX (sin x + i cos x)] = eX (cos x -- i sin x) + eX (sin x + i cos x)

= eX (cos x + sin x) + iex (cos x -- sin x).

Here we used the product rule with f(x) = e"', g(x) = sin x + i cos x.

(iii) D(x + iX 2)2 = 2(x + ix2 )(1 + i2x)

= 2(x -- 2x3) + i6x2

.

Complex functions have been introduced so we will have at our disposalthe complex exponential functions, erx

, with r complex. These functions,defined next, are the basic solutions of the linear homogeneous equation.

DEFINITION 3. If a and b are real numbers, then

e(a+ib)x = eax cos bx + ieax sin bx. (4)

4-1] COMPLEX-VALUED SOLUTIONS 85

The motivation for this definition is discussed in Problem 5. Let usnow compute the derivative of e(a+ib)x in accordance with (3).

De(a+ib)x = D[eax cos bx + ieax sin bx]

= aeax cos bx - beax sin bx + i(aeaX sin bx + beax cos bx)

(a + ib)eax cos bx + i(a + ib)eax sin bx

= (a + ib)[eax cos bx + ieax sin bx].

Comparing this formula with (4) we see that

whether r is real or complex.* (5)

Just as for real· functions, we say a complex-valued function is a solution of a differential equation if the equation becomes an identity whenthe function and its derivatives are substituted. For example, if Y = eix,then.y' = ieix, y" = _eix, and Y is a solution of y" + Y = O. We areof course primarily interested in real solutions, and the following theoremmakes the connection between real and complex solutions of homogeneouslinear equations.

THEOREM 2. If Y = u + iv, where u and v are real functions, then Y is asolution of

yen) + Pn_Iy(n-lJ + ... + PlY' + PoY = 0, (6)

where Po, ... ,Pn-l are real functions or real constants if and only ifu and v are solutions of (6).

Proof. The various derivatives of yare given by y(k) = U(k) + iv(k).Substituting y into the equation and separating the real and imaginaryparts, we get the complex function

[u(n) + Pn_IU(n-lJ + ... + PIU' + Po]

+ i[v(n) + Pn_IV(n-lJ + ... + PIV' + Pov].

This function is identically zero if and only if its real and imaginary partsare; i.e., if and only if both u and v are solutions of (6).

Returning to the exampl~ above, the fact that eix = cos x + i sin x isa solution of y" + y = 0 gives us the real solutions cos x and sin x.

N ow let us try to find solutions of the form y = erx for any homogeneouslinear equation with constant real coefficients.

yen) + Pn_Iy(n-lJ + ... PlY' + PaY = O. (7)

* One way of defining the real exponential eax is to say that it is the solutionof y' = ay such that yeO) = 1. The fact that eTX

, with r complex, satisfiesy' = ry, yeO) = 1 is one justification of the definition (4).


Substituting y = eTX, r possibly complex, we get

which is equivalent to

rn+Pn_Irn - 1 + ... + Pir + Po = O.

(8)

(9)

Thus eTX is a solution of (7) if r is a real or complex root of (9). As inSection 3-4, (9) will be called the auxiliary equation for (7). If (9) has acomplex root a + ib, then the conjugate complex number a - ib will alsobe a root (Problem 6), since the coefficients Pi are real. For each pair ofconjugate complex roots a ± ib of (9) we have the two complex solutionsof (7), eax cos bx + ieax sin bx and eax cos bx - ieax sin bx, and the two realsolutions eax cos bx and eax sin bx. For each real root r, we have the realsolution eTX

•

Therefore, if the auxiliary equation has no repeated roots, the required nreal solutions are the exponential functions corresponding to the real roots, andthe real and imaginary parts of the exponential functions corresponding to thecomplex roots. The proof that the n functions found in this way are linearlyindependent, and the solution of (7) when the auxiliary equation hasrepeated roots are given in Section 4-3.

EXAMPLE 2. y'" - 3y" + 7y' - 5y = 2 - 5x - 8e-x•

For this nonhomogeneous equation, we must find one particular solutionand the general solution of the reduced equation. The auxiliary equationis r3

- 3r2 + 7r - 5 = O. By inspection, 1 is found to be a root ofthe auxiliary equation, and therefore r - 1 is a factor of the left side.Division by r - 1 gives the other factor r2

- 2r + 5. The remainingtwo roots of the auxiliary equation are the roots of r2

- 2r + 5 = 0,which are found by the quadratic formula to be 1 + 2i and 1 - 2i. Thecomplex function eO +2i

)x is a solution of the reduced equation, and eX cos 2x,eX sin 2x are the corresponding real solutions. The general solution of thereduced equation is therefore

To find a particular solution of the given equation, we try a functionof the form A + Bx to fit the term 2 - 5x, and a function of the formCe-x to fit the term -8e-x , Substitution of y = A + Bx + Ce-x gives(7B - 5A) - 5Bx - 16Ce-x

. Therefore C = t, B = 1, and A = 1.The general solution is

4-1) COMPLEX-VALUED SOLUTIONS 87

PROBLEMS

1. Make a formal proof of Theorem 1 by filling in the details of this outline:First, suppose limx->xo u(x) = Uo and limx->xo v(x) = vo. Let E > O. Choosepositive numbers (h, (52 such that lu(x) - uol < E/V2 if 0 < Ix - xol < lhand Iv(x) - vol < e/V2 if 0 < Ix - xol < Oz. Show that If(x) - (uo + ivol < E

if 0 < Ix - xol < min (01, 02). Second, suppose limx->xof(x) = Uo + ivo. LetE > O. Pick 0 > 0 such that If(x) - (uo + ivo)/ < E if 0 < Ix - xol < O.Show that lu(x) - uol < E and Iv(x) - vol < E if 0 < Ix - xol < o.

2. Let f and g be differentiable complex functions with f = u + iv andg = r + is. Use the differentiation formulas for real functions and (3) to verifythe following:

(a) D(f + g) = Df + Dg (b) D(cf) = cDf, c a complex constant.(c) D(fg) = fDg + gDf (d) D(f/g) = (gDf - fDg)/g2

Note. Since the real functions are a subclass of the complex functions (they arethe complex functions u + iv with v = 0), these formulas hold if one function isreal and the other is complex.

3. (a) Differentiate eX sin x + iex cos x and check the result which was· ob. tained in Example l(ii) with the product rule.

(b) Expand (x + ix2)2 and then differentiate to check the chain rulecomputation of Example l(iii).

4. Compute the following derivatives.

(a) D(x sin x + i(x2 + 1)) (b) D(x2 + 2ix + 1 - ix2)(c) DV-1 - x2 (d) D[(x + ix2)(sin x + i cos x))

5. Show that the substitution of ib for x in the series for eX [(6), Section 2-6)gives eib = cos b + i sin b. The property ePe q = ep +q is an algebraic propertyof the series for eP, eq

, and ep +q and doesn't depend on whether P and q are realor complex. This shows that ea+ib = eaeib = ea(cos b + i ~in b), and e(a+ib)x =

eax+ibx = eax(cos bx + i sin bx).6. If z = a + ib, with a and b real, then the conjugate of z is the number

Z = a - ib. Let z = a + ib, and w = c + id be any complex numbers.

(a) Verify that z+ w = Z+ iii. (b) Verify that (zw) = ziii.(c) Let P(z) = Po + PIZ + ... + Pnzn, with Po, ... , pn real. Show that

P(z) = P(z).(d) Show that if Zo is a root of the polynomial equation P(z) = 0, P as in (c),

then zo is also a root.

7. Solve the following equations.(a) y(iv) - y = 4 + e2x (b) y'" + 3y" + 4y' + 2y = 2x

(c) y'" + 4y" + Sy' = 5(d) y'" - 6y" + 13y' - lOy lOx2 - 6x - 4

8. (a) Verify that 2 + 2i is a root of the equation

r4 - 5r3 + 40r - 96 = O.

(b) Solve the equation y(iv) - 5y'" + 40y' - 96y O.

88 SPECIAL METHODS FOR LINEAR EQUATIONS

ANSWERS

[CHAP. 4

4. (a) x cos x + sin x + i(2x)(b) 2x + i(2 - 2x)(c) ix(1 + xZ) -lIZ

(d) -x cos x + (xZ+ 1) sin x + i[(xZ+ 1) cos x + x sin xl7. (a) y = -4 + lsl1Z" + Cle" + cze-" + C3 cos X+ C4 sin x

(b) y = x - 2 + CIe-" + cze-" cos x + c3e-" sin x(c) y = x + CI + cze-z" cos x + c3e-z" sin x(d) y = -(x + 1)Z + eZ"(cl + Cz cos x + C3 sin x)

8. y = cle-3" + cze4" + c3ez" cos 2x + c4ez" sin 2x

4-2 Linear differential operators. An operator is a rule which associatesa unique function with each function in some set. An operator is thereforeitself a function, which is defined on some set of functions rather thannumbers, and whose values are functions. An operator A, like any otherfunction, can be defined by prescribing the value Ay of the operator at anarbitrary function y. Thus the formula Ay = y' + 3y defines the operatorA just as the formula z(x) = x Z + 2 defines the function z. For the Aand z just defined, for example, (Az)(x) = 2x + 3(xZ + 2).

The operators of interest to us. here, generalizations of the familiarderivative operators D, D Z

, etc., are those of the form

Ly = Pny(n) + Pn_ly(n-O + ... + PlY' + PoY (1)

or, equivalently,

where Po, PI, ... , Pn are constants. The operator L defined by (1) willalso be denoted by L(D) or by a formal polynomial in D,

L = L(D) = PnDn + Pn_lDn-l + ... + PID + Po. (2)

An operator of the form (2), as defined by (I), is called a linear differentialoperator. The general linear differential equation, with constant coefficients,can be written in operator notation as. follows:

Ly = L(D)y = q (Pn = 1), (3)

or

(Dn + Pn_IDn-l + ... + PID + Po)Y = q. (4)

The linearity properties of the operators (I) are expressed in the following theorem.

4-2] LINEAR DIFFERENTIAL OPERATORS 89

TIJEOREM 1. If L is a linear differential operator, then L(YI + Y2) =LYI + LY2' and L(cy) = cLy for all functions y, Yl, Y2 and all constants c.

Proof. The first property has already been proved, with different nota-tion, in Section 3-3 (the superposition principle). To show that L(cy) =eLy, note that PkDk(cy) = cPkDky, so c can be factored from each term inL(cy) to give eLy.

The advantage of the expression (2) lies in the fact that the ordinarymultiplication and addition of polynomials corresponds to the naturaloperator multiplication and addition defined below. This allows us tofactor the polynomial (2) and so replace a homogeneous linear equationwith equations of smaller order. Although we are interested in equations(4) with real coefficients, we may need to allow complex coefficients tofactor the operator (2). Since the rules for differentiation are the same forreal or complex functions and constants, it causes no difficulty to allowthe Pi in (2) to be complex and to admit complex-valued functions in thedefinition (1).

For any operators A and B, we define new operators A + Band AB by

(A + B)y = Ay + By,

(AB)y. A(By).

(5)

(6)

For example, if Ay = y' + 3y and By = xy, then (A + B)y = y' +3y + xy, and (AB)y = A(xy) = (xy)' + 3(xy) = xy' + y + 3xy. Notethat AB ~ BA in general; here, for instance, (BA)y = B(y' + 3y) =xy' + 3xy ~ (AB)y.

THEOREM 2. If L1 and L2 are linear differential operators, then so isL 1 + L 2 [defined in (5)] and the polynomial form (2) of L 1 + L 2 is theordinary formal sum of the polynomials for L 1 and L 2.

Proof. If

then

L 1 = PnDn + + PI D + Po,

L 2 = qmDm + + ql D + qo (say m < n),

(L 1 + L 2)y = L 1y + L 2y

= PnDny + ... + PI D y + PoY

+ qmDmy + ... + ql D y + qoY

= PnDny + ... + (Pm + qm)Dmy + ...+ (PI + ql)Dy + (PO + qo)Y.


This shows that L 1 + L 2 has the form (1) and is therefore a linear differential operator. Moreover, the polynomial form (2) of L 1 + L 2 is clearlythe sum of the polynomials for L 1 and L 2 •

Remark. The right side of (2) was introduced as a formal expression,with the plus signs serving only to make spaces between the terms. Byvirtue of Theorem 2 we can now regard the right side of (2) as a genuinesum, as defined in (5), of the operators Po, P1D, etc., and this is consistentwith the definition (1).

Recall that the coefficients in a linear operator are always assumed to beconstants. Theorem 2 would still hold if the coefficients Po, PI, ... , Pnwere allowed to be functions, but Theorem 3 would definitely not be truefor nonconstant coefficients.

THEOREM 3. If L 1 and L 2 are linear differential operators, then so isL 1L 2 and the polynomial form of L 1L 2 is the ordinary product of thepolynomials for L 1 and L 2 •

Proof. If L 1 and L 2 are as in Theorem 2, then

(L 1L 2 )y = L 1(L2y)

= L 1(f qjDjy)J=O

= EPiDi (~ qjDjy)

n m

= L Pi L qjDi+jyi=O j=O

n m

= L L PiqjDi+jy.i=O j=O

The riglit side above is of the form (1), showing that L 1L 2 is a lineardifferential operator. Also, the polynomial form of L 1L 2 is clearly theformal product

of the polynomials for L 1 and L 2 .

COROLLARY. If L 1 and L 2 are linear differential operators, then

L 1L 2 = L 2L 1•

Proof. Problem 3(c).The utility of the results above is illustrated in the following example.

4-2) LINEAR DIFFERENTIAL OPERATORS 91

EXAMPLE 1. Consider the homogeneous linear equation

y'll - y" + y' - y = 0.

This can be written in operator notation as

(D 3 - D 2 + D - I)y = 0.

The polynomial D 3- D 2 + D - 1 can be factored as

(7)

(8)

D 3- D 2 + D - 1 = (D 2 + I)(D - 1) = (D + i)(D - i)(D - 1).

Because of Theorem 3, we can write (8) in any of the equivalent forms:

(D 2 + I)[(D - I)y] = 0,

(D - I)(D + i)[(D - i)y] = 0,

(D - I)(D - i)[(D + i)y] = 0.

(9)

For any linear operator L, it is clear from (1) that LO = 0, where "0"denotes the zero function. It follows that y is a solution of (7), (8), or (9)if y is a solution of any of the equations

(D - I)y = 0,

(D - i)y = 0,

(D + i)y = 0.

(10)

Thus the third order homogeneous equation (7) is reduced to the threefirst order equations (10), whose solutions are e'X, eix = cos x + i sin x, ande-ix = cos x - i sin x. From Theorem 2, Section 4-1, we know that thecomplex functions are solutions of (7) only if their real and imaginaryparts are. So the required three real solutions of (7) are eX, sin x, and cos x.

EXAMPLE 2. (D 3- 2D 2 + 2D)y = 0.

This equation can be written in either of the forms

D[(D2- 2D + 2)y] = ° or (D 2

- 2D + 2)[Dy] = 0.

The solutions will be the solutions of Dy = 0, plus the solutions of(D 2

- 2D + 2)y = 0. Instead of factoring the quadratic operator intolinear factors, D 2

- 2D + 2 = [D - (1 + i)][D - (1 - i)], as inExample 1, we can use the method of Section 4-1. The roots of the auxiliary equation are 1 ± i, giving the real solutions eX cos x and eX sin x.

There is no real difference in the calculations involved in the methodsof Section 4-1 and the operator methods of this section as applied to


homogeneous linear equations with constant coefficients. Factoring thepolynomial operator certainly amounts to the same thing as solving theauxiliary equation. The operator methods, however, are the more flexibleand powerful. We will treat the case of repeated roots of the auxiliaryequation in the next section with operator methods. The next exampleillustrates the use of operators to solve a nonhomogeneous equation.

EXAMPLE 3. y" - y' - 2y = xex.

Let us write the equation in the form

(D - 2)[(D + l)y) = xex•

Clearly y will be a solution of (11) if and only if y is a solution of

(D + l)y = u,

where u is a solution of

(D - 2)u = xeX•

(11)

(12)

(13)

That is, we can replace the second order linear equations (11) by the pairof simultaneous linear first order equations (12) and (13). Multiplying(13) by the integrating factor e-2x, we get

and hence

ue-2x = Jxe-x dx + Cl·

Integration by parts gives

Jxe-x dx = -xe-x + Je-x dx

-e-X(x + 1),

and therefore the solutions u of (13) are·

u = -eX(x + 1) + cle2x. (14)

From (12) and (14), we see that y is a solution of (11) if and only if y is asolution of

This linear first order equation has the integrating factor eX, and we get

eXy' + eXy = -e2X(x + 1) + cle3x.

4-2] LINEAR DIFFERENTIAL OPERATORS 93

The solutions are given by

eXy = - f e2X(x + 1) dx + cle3x + Cz,

where cd3 has been replaced by Cl. Integration by parts gives

eXy = -teZX(x + 1) + f teZX dx + cle3x + Cz

-teZX(x + 1) + :i;e zx + cle3x + Cz.

The general solution of (11) is, accordingly,

y = -teX(2x + 1) + clezx + cze-x.

Note that the solutions cle zx + cze-x of the reduced equation appearautomatically, as they must since the procedure finds all solutions of (11).

EXAMPLE 4. (D - r)Zy = 0 (r is real.)

In Section 3-4 we introduced the solution of this equation and verifiedthat it worked. Let us now show how the method of Example 3 can beused to derive the solution. The equation is equivalent to the system

(D - r)u = 0,

The solutions of the first equation are

(D - r)y = u.

and hence the second equation becomes

Multiplying by the integrating factor e-rx, we get

e-rxy, - re-rxy = ClI

ye-rx = ClX + Cz,

y = erx(clx + cz).

PROBLEMS

1. Let A and B be the operators defined by Ay = yZ + xy, By 2y'.Compute the following:

(a) Axz (b) Bxz (c) (A + B)xZ

(d) A(BxZ) (e) B(AxZ)


2. Verify the superposition principle, L(Yl + Y2) = LYl + LY2, for lineardifferential operators L, using the polynomial notation for L (cL Theorem 3,Section 3-3).

3. (a) Prove that operator addition (5) is commutative and associative:A + B = B + A, and (A + B) + C = A + (B + C) for all operators A, B, C.

(b) Prove that operator multiplication is associative: (AB)C = A(BC) forall A, B, C.

(c) Prove the corollary to Theorem 3.

4. Solve the following equations by the method of Example 1.

(a.) (D3 - 6D2 + 5D)y = 0 (b) y,It + 2y" - y' - 2y = 0(c) (D2 - 1)(D2 + l)y = 0 (d) (D2 - 2D + 2)(D - l)y = 0

5. Show that if y is a solution of (D2 + l)y = e2x, then y is a solution of(D - 2)(D2 + l)y = O. Solve both equations.

6. Suppose that L is a linear differential operator and L(cos x) ;t. 0,L(sin x) ,t. O. Prove that the equation Ly = cos x has a solution of the formA cos x + B sin x.

7. Write an equivalent system of first order equations, as in Example 3, -andsolve (D - 1)2y = eX.

8. Write as a system and solve (D + l)(D - l)y = x.9. Solve (D - a)3y = 0 (a real) by changing to a system of three first order

equations.10. Recall from Section 3-4 that (D - a)2xeax = 0 (a real).

(a) Show that if y is a solution of (i) (D - l)(D ~ 2)y 2xex, then yis a solution of (ii) (D - 2)(D - 1)3y = O.

(b) Find all solutions of (ii) (see Problem 9).(c) Find all solutions of the reduced form of (i).(d) Find a particular solution of (i) by the method of undetermined

coefficients. Note that the form of your trial solution is determinedby your answers to (a), (b), and (c).

AJ.-"SWERS

1. (a) x4 + x3 (b) 4x (c) x4 + x3 + 4x(d) 20x2 (e) 8x3 + 6x2

4. (a) y = CI + c2ex + c3e5x

(b) y = clex + C2e-x + c3e-2x

(c) y = CleX+ C2e-X+ C3 cos X+ C4 sin x(d) y = clex + c2ex cos x + c3ex sin x

5. y = cle2x + C2 cos X+ C3 sin xy = ie2x + CI cos x + C2 sin x

7. y -!x2ex + eX(CIx + C2)8. y = -x + cle-x + c2ex

9. y = eax (clx 2 + C2X + C3)10. (a) y = CIe2x + eX(c2x2 + C3X + C4)

(c) y = CIe2x + c2ex

(d) y = (-x2 - 2x)eX

4-3] HOMOGENEOUS EQUATIONS: CONSTANT COEFFICIENTS 95

4-3 Homogeneous equations with constant coefficients. With themethods now at our disposal we can finish the discussion of the linearhomogeneous equation with constant real coefficients

(Dn + Pn_IDn-1 + ... + PID + PO)y = O. (1)

First let us recall from the preceding sections some general facts about (1).We know (Theorem 7, Section 3-3) that (1) has n linearly independent

solutions, and if YI, ... , Yn are any n linearly independent solutions, thenthe family

CIYI + .:.+ CnYn (2)

is the set of all solutions of (1).If f = u + iv is any complex valued solution of (1), then (Theorem 2,

Section 4-1) u and v are solutions of (1).The operator L = Dn + ... + PID + Po in (1) can be factored into

linear factors

(3)

where rI, r2, ... , r. are distinct numbers. Some of the numbers rj may becomplex, and if so, the complex rj occur in conjugate pairs (Problem 6,Section 4-1). All solutions of the equations

(4)

are (possibly complex) solutions of (1).The attack on (3) will proceed as follows. If rj is real, we will find k j

real solutions of (4) which are necessarily also solutions of (1). If rj iscomplex, then we find 2k j real solutions of

(5)

which are also solutions of (1). In this way we accumulate n real solutionsof (1). These solutions are linearly independent for any numbers rI, ... ,

r., and their linear combinations therefore constitute the general solutionof (1).

THEOREM 1. if m is a positive integer and r is a real or complex number,then

(6)

Proof. Calculating according to (5), Section 4-1, and the product rule,we see that

(D - r)xmer:· = D(xmerX ) - rxmerx

= mxm-Ierx + rxmerx _ rxmerx

= mxm-Ierx• (7)


Using (7) again with m replaced by m - 1 gives

(D - r)2xmeTX = (D - r)mxm-1eTx

= m(D - r)xm-1eTx

= m(m - l)xm - 2eTx . (8)

It is clear that we can repeat this process as long as the exponent of x ispositive, and after m steps, we get

Hence,

(9)

(10)

COROLLARY 1. If P is any of the numbers 0, 1, 2, ... , lc - 1, then(D - r)kxPeTX = O.

Proof. Problem 1.

COROLLARY 2. If P is any polynomial of degree k - 1 or less, then(D - r)kP(x)eTX = O.

Proof. Problem 1.

From Corollary 1, we see that if r is real, then

(11)

is a set of k real solutions of

(12)

If (D - r)k is a factor of the operator (3), then the solutions (11) of (12)are also solutions of (1). Thus we get k solutions of (1) corresponding toeach linear factor of multiplicity k in the operator.. Now suppose?, is complex, r = a + ib, and (D - r)k is a factor of(3). Then (D - 'F)k is also a factor ('F = a - ib). The equation

(D - r)k(D - 'Fly = 0 (13)

is a linear equation of order 2k with real coefficients (Problem 3). FromCorollary 1, this equation has the complex solutions

(14)

and also the solutions

(15)

4-3] HOMOGENEOUS EQUATIONS: CONSTANT COEFFICIENTS 97

The real and imaginary parts of the functions (14) are the same, apartfrom sign, as the real and imaginary parts of the functions (15); they arethe functions

eBX cos bx, xeBX cos bx, , xk-1eBX cos bx,

eBX sin bx, xeBX sin bx, , xk-1eBx sin bx.(16)

These solutions of (13) are of course also solutions of (1), and we havefound 2k real solutions of (1) corresponding to each pair of linear factors(D - r)k(D - r)k in the operator (3). Altogether we have found n realsolutions of the nth order equation (1).

THEOREM 2. Let Ly = 0 be an nth order linear equation with constantreal coefficients. Let

L = (D - rl)k1(D - r2)k2 (D - r.)k.,

where rl, ... , r. are distinct numbers.If rj is real, then Ly = 0 has the solutions

If rj = aj + ibj, then Ly = 0 has the solutions

eBjX cos bjx, xeBjX cos bjx, , xkj-1eBjx cos bjx,

eBjX sin bjx, xeBjX sin bjx, , xkj-1eBjx sin bjx.

The n solutions of Ly = 0 given above are linearly independent for anyoperator L, and the general solution of Ly = 0 is the set of all linear combinations of these n solutions.

Proof. The only statement in the theorem which has not yet been verified is the linear independence of the given solutions for any equation.The proof of linear independence in general involves such complicatednotation that it becomes uninstructive, and we will prove the statementonly in the case where the operator can be factored into real linear factors,

with al> ... , a. distinct real numbers. The linear combinations of thesolutions found for Ly = 0 can be written

(17)

where Pl>"" p. are arbitrary polynomials with respective degrees


k l - 1, ... , ks - 1. Let al be the largest of the distinct numbers aI,... , as, so that aj - al < 0 for j = 2, ... , s, and eCaj-a1)x is a decreasingfunction. Suppose that (17) is identically zero for some polynomialsPI, ... ,Ps• We must show that all the coefficients in each polynomial arezero, which is the same as showing that each polynomial is the zero function. If (17) is identically zero, then multiplying by e-a1x and transposingthe first term, we get

For any exponential eqX, q < 0, and any polynomial R(x), we have (Prob

lem4)lim R(x)eqX

= O.x-.oo

This implies, in view of (18), that limx-.oo Pl(x) = O. However, the onlypolynomial which tends to zero as x ~ 00 is the zero polynomial (Problem 5), so Pl(x) == O. Having shown that Pl(x) == 0, the assumption that(17) is identically zero becomes

The same argument as above can now be used to show that P 2 (x) == O.Continuing in this way, we see that (17) is identically zero only if all thecoefficients are zero, and the solutions found are linearly independent.

EXAMPLE 1.

(A) (D - 1)2(D2 - 2D + 2)y

= (D - 1)2[D - (1 + i)][D - (1 - i)]y = 0,

(B) (D + 2)(D2 - 4D + 13)2y

= (D + 2)[D - (2 + 3i)]2[D - (2 - 3i)]2y = 0,

y = cle-2x + C2e2x cos 3x + caxe2x cos 3x

+ C4e2x sin 3x + C5xe2x sin 3x

= cle-2x + e2x[(c2 + cax) cos 3x + (C4 + C5X) sin 3x].

(C) (D 2+ 4)2D2(D - l)y = (D + 2i)2(D - 2i)2D2(D - l)y = 0,

Y = (Cl + C2X) cos 2x + (ca + C4X) sin 2x

+ C5 + C6X + C7ex.

4-3] HOMOGENEOUS EQUATIONS; CONSTANT COEFFICIENTS 99

EXAMPLE 2. Let us show that the solutions eX, xex, eX cos x, eX sin x of

Example leA) are linearly independent. We could use the Wronskian test(corollary to Theorem 5, Section 3-3), but the computations involved areuninviting. Instead, we will argue directly from the definition. Supposethat for some numbers Cl, C2, Ca, C4, we have

Then

Cl + C2X + Ca cos x + C4 sin x == 0.

For x = 0,271",471", etc., this becomes

The linear function (Cl + ca) + C2X is zero for more than one x only if itis identically zero, so Cl + Ca = 0, and C2 = 0. For x = 71"/2, 71"/2 + 271",etc.,

Cl + Ox + Ca cos x + C4 sin x = Cl + £:4 = 0.

From Cl + Ca = 0, Cl + c~ = 0, we get Ca = C4, and the original assumption becomes

Cl + Ca (cos x + sin x) == 0.

This implies Cl = Ca = 0. Therefore, Cl = C2 = Ca = C4 = 0, and thesolutions are linearly independent.

PROBLEMS

1. Prove the corollaries of Theorem 1.2. Are the functions e", (x - l)e", (x2 - x)e" solutions of (D - l)ay = O?

Is this set of functions linearly independent?3. Show that (D - r)k(D - r)ky = [(D - r)(D - r)]ky = 0 is a linear

equation with real coefficients.4. Let q < O. Use I'Hospital's rule to show that lim,,->co xeq

" = O. [Hint:xeq" = x/e-q,,]. Prove by induction that lim,,->co xneq" = 0 for any positiveinteger n. Show that lim,,->co R(x)eq" = 0 for any polynomial R.

5.·Show that if R is any nonconstant polynomial, then lim,,->co IR(x)1 = 00.

[Hint: If x > 0, then


Show that if x is sufficiently large and pn ¢ 0, then the second factor on the rightis greater than IPnl!2, and IR(x)1 > iIPnlxn.]


(a) (D - 1)2(D2 - 2D + 2)y = 0(b) (D2 + 1)(D2 + 4D + 5)2y = 0(c) (D2 - 1)(D2 + 5D + 4)2y = 0(d) (D2 - 4)(D + 2)(D2 + 4)y = 0

(e) (D6 - l)y = 0 (f) (Da + D2 + iD)y = 0(g) (D4 - D2)y = 0 (h) (D4+ 2Da+ 2D2+ 2D+ l)y = 0

7. Show that the solutions are linearly independent.

(a) solutions of Problem 6(a) (b) solutions of Problem 6(b)

8. Suppose Yo is a solution of Ly = 0, where L is a linear differential operator(with constant coefficients). Prove that the following functions are necessarilyalso solutions of Ly = 0 or give an example (of L and yo) to show they are not.

(a) yff - 2 (b) 3Yb + 2yo

ANSWERS

(c) xyo

6. (a) y = (CI + c2x)e" + cae" cos x + c4e" sin x(b) y = CI cos x + C2 sin x + e-2"[(ca + C4X) cos x + (cs + C6X) sin x](c) y = CIe" + (C2 + cax + c4x2)e-" + (cs + c6x)e-4"(d) y = cle2" + (C2 + cax)e-2" + C4 cos 2x + cs sin 2x

() "+ -"+ 0/2),,( V3 + . V3 )e y = cle c2e e ca cos 2 x C4 sm 2 x

+ -(112),,( V3 + . V3 )e .cs cos 2 X Cs sm 2 x

+ -(112)" ( X . X)(f) Y = CI e C2 cos 2+ Ca sm 2(g) y = Cl + C2X + cae" + c4e-"(h) y = (CI + c2x)e-" + ca cos x + C4 sin x

8. (b) L(3D + 2)yo = (3D + 2)Lyo = (3D + 2)0(a) and (c) are not necessarily also solutions.

4-4 Method of undetermined coefficients. In the preceding section wesaw that the solutions of

and

(a real)

(r = a + ib)

(1)

(2)

are the functions

P(x)eax cos bx + Q(x)eax sin bx, (3)

4-4] METHOD OF UNDETERMINED COEFFICIENTS 101

where P and Q are any polynomials of degree k - 1 and b = 0 in case (1).Since any linear operator can be factored into operators such as those in(1) or (2), it follows that all solutions of a linear equation Ly = 0 withconstant coefficients are sums of functions of the form (3). Now look atthese facts in this way: For any function q which is of the form (3), thereis a linear operator L such that Lq = O. This observation is the centralidea in the following theorem.

THEOREM 1. If L is a linear differential operator and P, Q are any givenpolynomials, then the equation

Ly = P(x)eaX cos bx + Q(x)eaX sin bx

has a particular solution Yo which can be written

(4)

(5)Yo = P*(x)eaX cos bx + Q*(x)eaX sin bx

for some polynomials p* and Q*.

Proof. We give the proof for the case b ~ 0, and ask the student tosupply the proof for the case b = 0 (Problem 10).

Let k - 1 be the larger of the degrees of P and Q. Then by Corollary 2of Theorem 1, Section 4-3,

[D - (a + ib)]k[D - (a - ib)]k(P(x)eaX cos bx + Q(x)eaX sin bx) = O.

It follows that every solution of (4) is a solution of

[D - (a + ib)]k[D - (a - ib)]kLy = O. (6)

We know what all the solutions of (6) are, so we know what form a solutionof (4) must take. Since we are looking for a particular solution of (4), wecan disregard solutions of (6) which are solutions of the reduced form of(4) (i.e., of Ly = 0). The solutions of (6) which are not solutions ofLy = 0 are among the solutions corresponding to the factors

[D - (a + ib)]k+i[D - (a - ib)]k+i, (7)

where we assume that [D - (a + ib)] occurs in L with multiplicity j(j possibly zero). The solutions of (6) corresponding to the factors (7) arethe functions

P*(x)eaX cos bx + Q*(x)eaX sin bx, (8)

where p* and Q* are arbitrary polynomials of degree k + j - 1. Sinceall solutions of (4) which are not solutions of Ly = 0 appear among thefunctions (8), there will be some specific polynomials p* and Q* for which(8) is a solution of (4). If [D - (a + ib)] does occur as a factor of L(i.e., j ~ 0), then some of the terms in (8) will be solutions of Ly = O.


We can ignore these and take p* and Q* of the form

EXAMPLE 1.

(D 3- D 2 + D - l)y = (D - 1)(D2 + l)y = xex . (9)

Since (D - 1)2xeX = 0, all solutions of (9) are solutions of

Some particular solution of (9) therefore must be a solution of (D - 1)3y =O. The solutions of this, omitting cex which is a solution of the reducedform of (9), are the functions

y = (Aa: + Bx2)ex•

The derivatives of the functions (10) are

y = [Ax + Bx 2]eX,

Dy = [A + (A + 2B)x + Bx2]e"',

D 2y = [(2A + 2B) + (A + 4B)x + Bx2]ex,

D 3y = [(3A + 6B) + (A + 6B)x + Bx2]ex•

(10)

Substitution in (9) gives

[(2A + 4B) + 4Bx + Ox2]ex = xex•

Therefore (10) is a solution of (9) if 4B = 1 and 2A + 4B = 0; that is,if B = ! and A = -!. The general solution of (9) is

y = (Cl - !x + !x2)ex + C2 cos X + C3 sin x.

EXAMPLE 2.

(D 4 - D 3 + 4D2- 4D)y = D(D - 1)(D2 + 4)y = x + cos 2x.

(11)Since D 2x = 0, and (D 2 + 4) cos 2x = 0, we have

D 2 (D 2 + 4)(x + cos 2x) = O.

Therefore all solutions of (11) are solutions of

(12)

4-4) METHOD OF UNDETERMINED COEFFICIENTS 103

The solutions of (12) which are not solutions of the reduced form of (11)are

y = Ax + Bx2 + Cx cos 2x + Ex sin 2x. (13)

Therefore, for .some numbers A, B, C, E, determined by substitution in(11), (13) will be a particular solution of (11).

The following theorem can frequently be used to simplify the computations involved in the method of undetermined coefficients.

THEOREM 2. If L = L(D) is a linear differential operator (with constantcoefficients) and F is any sufficiently differentiable function, then

Proof. Let L(D) be the operator

L(D) = PnDn + ... + PID + po.

(14)

(15)

By L(D + r) we mean the operator obtained by replacing D in (15) by(D + r); that is,

L(D + r) = Pn(D + r)n + ... + PleD + r) + Po. (16)

Let us first calculate DerxF(x), D 2erxF(x), etc.

DerXF(x) = erxDF(x) + rerxF(x)

= erx(D + r)F(x).

Using (17) again we get

DVXF(x) = D[erx(D + r)F(x)]

= erx(D + r)2F(x).

Proceeding in this way, we find that

for all k. Therefore

(17)

(18)

(19)

L(D)erXF(x) = [PnDn + ... + PID + po]erxF(x)

= PnDnerxF(x) + ... + PIDerxF(x) + poerxF(x)

= Pnerx(D + r)nF(x) + ... + Plerx(D + r)F(x) + poerxF(x)

= erx[Pn(D + r)n + ... + PI (D + r) + Po]F(x)

= erxL(D + r)F(x).


COROLLARY. L(D)eTX = eTXL(r).

Here L(r) is the number obtained by substituting r for Din (15) i that is,

L(r) = Pnrn + ... + Pir + Po.

The proof of the corollary is left to the student (Problem 11).

EXAMPLE 3. Let us use Theorem 2 to perform the calculations ofExample 1. We know there is a solution of (D - 1)(D2 + l)y = xex ofthe form (Ax + Bx2)ex. The substitution can be effected as follows:

(D - 1)(D2 + l)eX(Ax + Bx2)

= eX[(D + 1) - l][(D + 1)2 + l)(Ax + Bx2)

= eXD(D2 + 2D + 2) (Ax + Bx2)

= eX(D2 + 2D + 2)(A + 2Bx)

= eX(4B + 2A + 4Bx).

EXAMPLE 4. Find a particular solution of D 2(D - 1)(D2+ 2)y = 3e2x.By Theorem 1 there is a solution of the form Ae2x, and by the corollaryabove,

D 2(D - 1)(D2 + 2)Ae2X = AD2(D - 1)(D2 + 2)e2X

= Ae2X22(2 - 1)(22 + 2) = 24Ae2x.

Therefore ie2x is a particular solution.

PROBLEMS

Solve the following equati<ms.

1. (D3 - D)y = 2e" 2. (D2 - 3D + 2)y = cosx3. (D4 - l)y = xe" 4. (D3 + DZ - 2D)y = XZ

5. (D - 2)Z(D - l)y = 4e2" 6. (DZ + 4)(DZ + l)y = 5 sin x

7. (DZ + 1)(D - 1)(D - 2)y = (XZ+ l)e"8. (D - 1)(DZ - 2D + 2)y = e" cos x9. (D - 2)Z(D + l)y = 18xez"

10. Prove Theorem 1 for the case b = O.11. Prove the corollary to Theorem 2.

ANSWERS

1. Y = xe" + C1 + c2e" + c3e-"2. y = lo cos x - 1

30 sin x + cIe" + c2e2"

3. y = (-jx + txZ)e" + cIe" + c2e-" + C3 cos X + C4 sin x

4-5] INVERSE OPERATORS 105

4. Y =-lx - !x2 - tx3 + C1 + c2ex + C3C2x

5. Y = 2x2e2x + clex + (C2 + c3x)e2x

6. Y = -ix cos x + Cl cos X+ C2 sin x + C3 cos 2x + C4 sin 2x7. Y = (-x - tx3)ex + C1 cos x + C2 sin x + c3ex + C4e2x

8. Y = -!xex cos x + clex + c2ex cos x + c3ex sin x9. y = (-x2 + x3)e2x + (Cl + c2x)e2x + c3e-x

4-5 Inverse operators. There are many manipulations with operatorswhich simplify the process of finding particular solutions for linear equations. We will look at a few techniques associated with the idea of aninverse operator. One inverse operator, the indefinite integral, is familiarfrom calculus. The notation ff(x) dx = F(x) is used to indicate thatDxF(x) = f(x); in this sense, f( ) dx is inverse to Dx. We extend this ideaand describe an inverse for any linear differential operator L(D).

AGREEMENT. We will write y = [l/L(D)]q(x) or y = L(D)-lq(X) ifL(D)y = q(x). That is, we write y = L(D)-lq(X) to indicate that y issome particular solution of L(D)y = q(x).

We define the sum and product of two inverse operators in the naturalway,

[Ll(D)-l + L 2(D)-l]q(X) = Ll(D)-lq(X) + L 2(D)-lq(X), (1)

[L l (D)- lL 2(D)-l]q(X) = L l (D)-l[L2(D)-lq(X)]. (2)

The following formulas, which express the linear properties of inverseoperators, follow directly from Theorem 1 of Section 4-2.

L(D)-l[ql(X) + q2(X)] = L(D)-lql(X) + L(D)-lq2(X), (3)

L(D)-l[cq(X)] = cL(D)-lq(X). (4)

The commutative property (corollary to Theorem 3, Section 4-2) oflinear differential operators, and the definition (2) above yield the formula

[L l (D)L2(D)]-lq(X) = [L2(D)-lL l (D)-l]q(X)

= [L l (D)-IL2(D)-l]q(X). (5)

Recall from Section 2-4 that the first order linear equation

(D - a)y = q(x)

has a particular solution

(6)

(7)


This fact can be written

Y = D ~ a q(x) = (D - a)-lq(x) = eaxfe-aXq(x) dx. (8)

From (8) and (5) it follows that a particular solution of

(D - a)(D - b)y = q(x)

can be writteny = (D - a)-I[(D - b)-lq(X)]

= eaxfe-a"[ebX f e-bXq(x) dxJ dx.

(9)

(10)

The method indicated in (10) of expressing a particular solution in termsof repeated integrals can also be extended to equations of higher order.

EXAMPLE 1. (D - 2)(D - I)2y = eX.

A particular solution y can be obtained as follows:

y = (D - 2)-I(D - I)-I[(D - I)-leX]

(D - 2)-I(D - I)-I[eXf e-xex dx]

(D - 2)-I[(D - I)-I(xeX)]

(D - 2)-I[eXfe-xxex dx]

(D - 2)-I[tx2eX]

= e2x f e-2Xtx2eXdx

te2Xf x2e-x dx

-teX (x 2 + 2x + 2).

If the inverse operators are applied in a different order, one gets a differentparticular solution; for example (Problem 1),

These two solutions differ by -tex, which is a solution of the reduced

equation (D - I)2(D - 2)y = O.Note that the process illustrated in the example really amounts to

considering the system(D - I)u = eX,

(D - I)v = u,

(D - 2)y = v,

4-5] INVERSE OPERATORS 107

(11)A + BD-a D-b

1(D - a)(D - b)

as we did in Examples 3 and 4 of Section 4-2, except here we ask for onlyone solution of each equation.

The method of Example 1 can be used to find a particular solution ofany equation L(D)y --.:. q(x), even if some of the linear factors of L(D) arecomplex. In most cases, however, the integrations involved become soinvolved that the method is not practical. The following theorem providesa method which usually leads to simpler calculations than (10).

THEOREM 1 (Partial fractions method). If a and b are distinct realnumbers, then the following equation is an operator identity if 1:t is a formalalgebraic idenl1"ty:

Proof. The assumption is that A and B are the numbers such that·A(D - b) + B(D - a) = 1. The assertion is that AYl + BY2 is asolution of (D - a)(D - b)y = q(x) if Yl = (D - a)-lq(x) andY2 = (D - b)-lq(X). Let us assume, therefore, that Yl and Y2 areparticular functions satisfying (D - a)Yl = q(x), and (D -'- b)Y2 -:- q(x).Then

(D - a)(D - b)[AYl + BY2]

= A(D - b)[(D - a)yd + B(D - a)[(D - b)Y2]

= A(D - b)q(x) + B(D - a)q(x)

= [A(D - b) + B(D - a»)q(x)

= [I)q(x) = q(x).

EXAMPLE 2. (D + 1)(D - 2)y = xex.

1 x [-i i] xY = (D + l)(D - 2) xe = D + 1 + D - 2 xe

1 1 x+l 1 x-3D+l xe 3D_2 xe

-ie-xJeXxex dx + ie2xJe-2xxex dx

-ie-X[e 2X(!x - !») + ie2X[e-X(-x - 1»)

-neX(2x - 1) - neX(4x + 4)

= -!eX (2x + 1).

By way of review, let us check this last answer using (9) of Section 4-4.

(D + l)(D - 2)[-!eX(2x + 1»)

-!eX(D + 2)(D - 1)(2x + 1)

-!ex(D + 2)(2 - 2x - 1)

_!eX(-2 - 4x + 2) = xex.


Theorem 1 extends to third and higher order operators. For example,if a, b, and c are distinct numbers, and A, B, and C are the numbers suchthat

1(D - a)(D - b)(D - c)

is an algebraic identity, then

A + B + _C_ (12)D-a D-b D-c

is a particular solution of (D - a)(D - b)(D - c)y = q(x) (Problem 3).In the last sections we have derived the formulas

L(D)eTXF(x) = eTXL(D + r)F(x),

L(D)eTX = eTXL(r),

(D - r)kxkeTX = k!eTX .

The following formulas are simply restatements of those above in terms ofinverse operators.

1 TXF(') TX 1 F( )L(D) e x = e L(D + r) x,

1 TX eTXL(D) e = L(r) , (if L(r) ~ 0),

1 TX xkeTX

(D - r)k e = ----,cr.

EXAMPLE 3. (D - 2)3(D + 2)2(D - l)y = e2x.

A particular solution is given by

1 [ 1 ] 2xY = (D - 2)3 (D + 2)2(D - 1) e

1 [e 2X]= (D - 2)3 16 [from (14)]

1 x 3e2x 1 3 2x= 16 3! = 96 x e [from (15)].

PROBLEMS

(13)

(14)

(15)

1. Find a solution of the equation of Example 1 by evaluating the inverseoperators in this order: (D - I)-I(D - 2)-I(D - I)-lex.

2. Use (10) to find a solution of (D + I)2y = e-X In Ixl and check youranswer.

4-6] VARIATION OF PARAMETERS METHOD 109

3. Show that (12) is an operator identity if it is an algebraic identity (comparethe proof of Theorem 1).

4. Use the partial fractions method (12) to solve

(D + 1)(D - I)(D - 2)y = eX.

5. Find a particular solution of (D - I)(D + 2)y = e-2x and use (14) ofSection 4-4 to check your answer.

(a) use formula (10) (b) use formula (13)

6. Use (14) and (15) to solve the following:

(a) (D - 2)2(D + l)y = e2x (b) D(D + I)(D - 2)2y = e-x.

7. Let r be a complex number and 'i' its conjugate. Show that formula (10)can be used to find the real solutions

1 1y = -= x + -_- (r + r)rr (rr) 2

of (D - r)(D - r)y = x.

8. Use the formula of Problem 7 to solve (D2 - 2D + 5)y = x.9. Use Theorem 1 to find a particuiar solution of (D2 - 3D + 2)y = sin2x.

[Hint: !eBX

sin2

xdx = a2e: 4 [asin2

x - 2sinxcosx+~J.J

ANSWERS

1. y = -!ex(x2 + 2x)2. y = ix2e-X (2In Ixl - 3)4. Y = -ieX (1 + 2x)5. (a) and (b) y = -le-2x(3x + 1)6. (a) y' = ix2e2x (b) y = -lxe-x

8. y = !x + 225

9. y = -lo sin2 x + 230 sin x cos x + l~

4-6 Variation of parameters method. In this section we give a methodfor finding a particular solution of the linear equation

yen) + ... + PlY' + PoY = q (1)

whenever we have n linearly independent solutions of the reduced equation.Here we are not assuming that the coefficients Po, PI, ... , Pn-l are constants, but that these are arbitrary functions continuous on some commoninterval. Since our concern so far has been primarily with equations withconstant coefficients, let us recall that the theory of Section 3-3 applies toany linear equation, whether the coefficients are constants or functions.


Thus to solve (1), we must find n linearly independent solutions of thereduced equation and one solution of (1). Although we give no generalmethods for solving the reduced equation if the coefficients are not constant, the variation of parameters method insures that if we can solve thereduced equation, then we can solve (1) for any right-hand member q.

Let us first illustrate the method for the second order equation.

y" + PlY' + PoY = q. (2)

The assumption is that we have two linearly independent solutions YI, Y2of the reduced equation

y" + PlY' + PoY = O.

We try to find a solution Y of (2) in the form

(3)

(4)

where VI and V2 are functions to be determined. In general, there are manyfunctions VI, V2 such that VIYI + V2Y2 is a solution of (2) (see Problem 1).Since it is reasonable to assume that we can put additional conditions onVI and V2, we make an assumption which facilitates the computations.We show that there are functions VI and V2 such that the functionY = VIYl + V2Y2 is a solution of (2) and also satisfies the equation

(5)

This last assumption is clearly equivalent to the condition

Computing y" from (5), we get

y" = VIY'{ + V2Y'{ + yivi + Y~v~. (6)

Now we substitute Y, y', and y" as given by (4), (5), and (6) in Eq. (2)and group the terms containing VI and those containing V2. The result is

The terms in brackets are zero, since YI and Y2 are solutions of (3), and thecondition that Y be a solution of (2) is therefore

yivi + Y~v~ = q.

That is, the function Y of (4) will have the derivative (5) and be a solution

4-6] VARIATION OF PARAMETERS METHOD 111

(8)

of (2), if VI and V2 satisfy the equations

Ylvi + Y2V~ = 0,

Yivi + y~v~ = q.

For each x, the equations (8) are simultaneous linear equations in v{(x)and v~(x). The determinant of this system is the Wronskian of the twoknown functions YI and Y2:

W(YI(X), Y2(X)) = IYI(X) Y2(X) I·yi(x) y~(x)

Since YI and Y2 are linearly independent solutions of the reduced equation,the Wronskian is never zero, and the equations (8) have the solutions

viex) = -Y2(X)q(X)/W(YI(X), Y2(X)),

v~(x) = YI(X)q(X)/W(YI(X), Y2(X)).(9)

Since the functions on the right of equations (9) are continuous, they haveantiderivatives, and there are functions VI and V2 satisfying (9) and hence(8). For these functions, VIYI + V2Y2 is a particular solution of (2), andthe general solution of (2) is

Y = (CI + VI)YI + (C2 + V2)Y2.

EXAMPLE 1. y" - Y = e2"'.

Two linearly independent solutions of y" - Y = 0 are e'" and e-"'. LetY = VIe'" + V2e-"', and suppose that y' = VIe'" - V2e-"'; that is, supposethat e"'v{ + e-"'v~ = O. Then

Substitution in the differential equation gives

In other words, Y will have the given derivative, and be a solution, if

and

Solving these equations, we get v{ = !e"', VI = !e"', and v~ = -!e3"',V2 = -i-e3"'. The equation therefore has a particular solution


EXAMPLE 2. Y" + Y = sec x.

Two solutions of the reduced equation are cos x and sin x. We set

Y = VI cos X + V2 sin x,

y' = -VI sin x + V2 cos x,

y" = -VI cos X - V2 sin x - vi sin x + v~ cos x.

Then y will be a solution with the derivatives given above if

vi cos x + v~ sin x = 0,

-vi sin x + v~ cos x = sec x.

The determinant of this system is W(cos x, sin x) = 1, and solving, we get

, 10 sin xl sin xV I = sec x cos x = - cos x '

V' = I cos x ° I= 12 -sin x sec x .

Therefore VI = lnlcos xl, V2 = x, and a particular solution of the equationis

y = cos x lnlcos xl + x sin x.

N ow we will generalize these ideas so that they apply to a linear equation of any order.

THEOREM 1. If YI, ... , Yn are linearly independent solutions of thereduced form of

y(n) + Pn_Iy(n-ll + ... + PlY' + PoY = q,

then there are functions VI, . . . , Vn which satisfy

(10)

+ + Ynv~

+ + Y~v~

= 0,

= 0,

(11)

yin- 2)vi + + y~n-2)v~ = 0,

yin-l)vi + + y~n-l)v~ = q.

and for any such functions VI, .•• , Vn, thefunctiony = VlYI + ... + VnYnis a solution of (10).

Proof. The system (11) of linear equations can be solved for v{, ... , v~,

since the determinant of the system is W(YI' ... ,Yn), which never van-

4-6] VARIATION OF PARAMETERS METHODS 113

ishes.Moreover, the solutions V{, ... , v~ can be written as quotients ofdeterminants whose entries are continuous functions. Hence the expressions for V{, ... , v~ will be continuous, and will have antiderivativesVI, ... ,Vn. Therefore, there are functions satisfying (11).

The first (n - 1) of the equations (11) are the conditions that the first(n - 1) derivatives of yare those given below; the nth derivative of y issimply calculated from y(n-ll without further assumptions.

+ ... + VnYn,

(12)y' = VlYS. + + vnY~,

y(n-l) = vlyin- l ) + + vny~n~1),

y(n) = vlyin) + + vny~n) + yin-l)vi + ... + y~n-l)v~.

Now substitute y, y', ... , y(n) from (12) into equation (10). The coefficientof Vl after substitution is the first column of the array (12) with the appropriate coefficients; namely,

(13)

Since Yl is a solution of the reduced form of (10), the coefficient (13) of Vlis zero. Similarly, the terms containing V2, ... , Vn drop out, since Y2, ... ,Yn are solutions of the reduced equation. The condition that y be a solution of (10) is therefore simply

This equation is satisfied by the last of the assumptions (11), and hencey is a solution of (10).

EXAMPLE 3. y'" - 3y" + 2y' = e-x.

The functions 1, eX, e2x are linearly independent solutions of the reducedequation. We find the functions VI, V2, and V3 such that if

theny' = V2ex + 2v3e2x

y" = V2ex + 4v3e2x

(that is, vi + eXv~ + e2zv3 = 0),

(that is, eXv~ + 2e2Xv3 = 0),

andy is a solution

114 SPECIAL METHODS FOR LINEAR EQUATIONS [CRAP. 4

Similarly, v~ = -2ex/(2e 3x) = -e-2X, and Va = 1/(2e3X) = !e-3x. Integration gives VI = -!e-X, V2 = !e-2x, and V3 = -i-e-3x. A particularsolution of the equation is therefore

PROBLEMS

1. Find functions VI and V2 such that vI(x)eX+ v2(x)e2x is a solution ofy" - 3y' + 2y = 4x - 4, and such that

(a) VI (x) = 0 (b) VI (X)YI (x) = V2(X)Y2(X)(c) VI and V2 satisfy (10).

2. Write out in detail the statement and proof of Theorem 1 for the case n = 3.3. Solve by variation of parameters: y" - y = xeX

•

4. Solve the equation y" + y = csc x.5. Check that eX is a solution of (1 - x)y" + xy' - y = O. Find a second

(polynomial) solution of the reduced equation and solve (1 - x)y" + xy' - y =(1 - x)2. [Caution: Do the equations (9) and (11) apply directly to the givenequation in its present form?]

6. Show that x and l/x are solutions of y" + (l/x)y' - (l/x 2)y = 0: Solveby variation of parameters: y" + (l/x)y' - (l/x2)y = In x (x > 0).

7. Solve by variation of parameters: y'" - y" = x3 •

8. Solve by variation of parameters: y'" - 2y" - y' + 2y = eX.9. If Yo is one solution of the reduced form of a linear equation, then the sub

stitution y = YOu gives a linear equation in u in which u itself is missing. Thusthe order is effectively reduced by one (substitute V = u'). Illustrate this forthe second order equation y" + PlY' + POy = q.

10. Solve by the method of Problem 9.

(a) Problem 5, using Yo = ex; (b) Problem 6, using yo = x.

11. Use the substitution of Problem 9 to solve (x 2 - x)y' + (1 - 2x)y-x2, after checking that Yo = x 2 - x is a solution of the reduced form.

4-6) VARIATION OF PARAMETERS METHOD

ANSWERS

115

1. (a) V2(X) = (2x + 1)e-2"(b) VI(X) = (x + !)e-"(c) VI(X) = 4xe~", V2(X) = -(2x - l)e-2"

3. Y = cle" + C2e-" + -ke"(2x2 - 2x + 1)4. Y = (CI - x) cos x + (C2 + In Isin xl) sin x5. Y = cle" + C2X + 1 + x + x 2

6. VI = !x (In x-I), V2 = -tx3 In x + lsx3,Y = CIX + C2/X + kx2 In x - t-x2

7. VI = ix5

- tx4, V2 = -tx4, V3 = e-"(-x3 - 3x2 - 6x _ 6),

Y = -iox5- tx4

- x 3 - 3x2 - 6x - 6 + CI + C2X + c3e"8. VI = -!x, V2 = l2e2", V3 = -!e-",

Y = cle" + C2e-" + c3e2" - te"(2x + 1)11. Y = x + c(x2 - x)

CHAPTER 5

THE LAPLACE TRANSFORM

5-1 Review of improper integrals. The Laplace transform is definedin terms of an improper integral, and we will review here some of the factswe will need about such integrals.

Recall that the Riemann integral

lab f(x) dx

is initially defined only for finite intervals [a, b], and only for functions fwhich are bounded on the given interval. This definition is then extendedto the so-called improper integrals, in which the integrand is unboundedon some finite interval, or the interval of integration is infinite. We willbe concerned only with integrals which are improper because the upperlimit of integration is infinite. Specifically, we consider integrals of theform

(1)

where f is continuous on [0, (0). The integral (1) is defined as the limit

(" f(x) dx = lim (R f(x) dx.Jo R-+", Jo

(2)

If this limit exists, we say the integral (1) converges, otherwise we say theintegral diverges. The following comparison test, similar to that for series,is a basic method for showing the convergence of (1).

THEOREM 1. If If(x) I ~ g(x) for all suffidently large x, and f~ g(x) dxconverges, then f~ f(x) dx converges.

Notice that, in particular, Theorem 1 says that f~ f(x) dx converges iff~ If(x) Idx converges.

Now consider the case in which the integrand contains a parameter.Suppose f(s, x) is continuous for a ~ S ~ {3 and x ~ 0 and that thefollowing integral converges for each s in [a, (3].

10'" f(s, x) dx = Ij!(s).

116

(3)

5-1]

If we write

REVIEW OF IMPROPER INTEGRALS

RF R(S) = fo f(s, x) dx,

117

(4)

then (3) is the same as saying that for each s in [a, til

lim FR(S) = ~(s).R---+oo

(5)

That is, for each S in [a, mand every e > 0 there is a numb~rN (dependingon s and e) such that

(6)

whenever R ;::: N. The situation in (5), or (3), is similar to the convergenceof a sequence of functions {Fn I, except here we have a function FR for eachnumber R ;::: O.

The function ~ defined in (3) need not be continuous unless someassumption is made about how the convergence of the integral depends on s.We say the integral (3) converges uniformly for s in [a, til if for every e > 0there is a number N (depending only on e) such that IFR(S) - ~(s)1 < efor all s in [a, mif R ;::: N.

THEOREM 2. If the integral (3) converges uniformly for s in [a, til, then~ is continuous on [a, m.The following theorem provides a convenient test for uniform con

vergence.

THEOREM 3 (Weierstrass M-test for integrals). If M is a continuousfunction on [0, (0) such that J~ M(x) dx converges, and If(s, x)1 ~ M(x)for all s in [a, mand all x ;::: 0, then J~ f(s, x) dx converges umformly on[a,tll.

Uniform convergence is also the critical hypothesis for showing that thefunction ~ of (3) is differentiable, and that ~/(S) can be found by differentiating under the integral sign in (3).

THEOREM 4. If f.(s, x) is continuous for s in [a, mand x ;::: 0, and theintegral

converges umformly on [a, til, then ~ is differentiable and

~/(S) = 1000

f.(s, x) dx.

118 THE LAPLACE TRANSFORM [CHAP. 5

Note that it is the integral of the partial derivative f8(s, x) which mustconverge uniformly.

EXAMPLE 1. Find the values of s for which the integral

(7)

converges, and find a formula for cf>(s).

For s = 0, the integrand is identically one, and the integral diverges.If s r! 0, then

r~ (1 + s2)e-8X dx = lim (1 + S2) rRe-8X dx

Jo R->oo Jo

= lim _ (1 + S2) (e-8R _ 1).R->oo S

This last limit exists if s > 0 (limR->oo e-8R = 0), and fails to exist if s < 0(limR->oo e-8R = 00). Therefore the integral (7) converges if and only ifs > 0, and for these values of s

cf>(s) = (1 + S2) .S

EXAMPLE 2. Show that the function cf> defined by

cf>(s) = 10'" e-8X dx

is defined and differentiable for s > O.

(8)

(9)

As in the example above it is easy to show that the integral in (8) converges if and only if s > O. Since (a/as)e-8X = -xe-8X

, we want to showthat

cf>' (s) = 1000

_xe-8X dx

for any s > O. To verify (9), we show that the integral of (9) convergesuniformly on some interval [a,,8] around s, for each s > O. For any numbers a and ,8 with 0 < a < ,8 and any number s in [a, ,8], we have

I -8xl -8X < -ax-xe = xe _ xe ,

and the integral(10)

converges (Problem 1). By Theorem 3, with M(x) .= xe-ax, the integral

5-2) THE LAPLACE TRANSFORM 119

in (9) converges uniformly on [a, f3]. It then follows from Theorem 4 that<p'(s) exists and is given by (9) for all s in [a, f3]. Since any number s > 0is in some interval [a, f1] with a > 0, et/(s) exists for all s > 0.

PROBLEMS

1. Show that (10) converges, given that a > O.2. Evaluate the integrals in (8) and (9), and verify that et/(s) is given by (9).3. Let <p(s) = Jo'" xe-'''' dx. Show that <p(s) is defined for s > 0 and that

<p'(s) exists for s > O.4. Let p be a fixed number with 0 < p < 1. Verify that the following integral

converges uniformly for s in [p, 1):

1'" sin (sx) -'" d---e x.o s

5. Show that the integral of Problem 4 can be differentiated under the integralsign for sin [p, 1).

5-2 The Laplace transform. Suppose that f is a function on [0, (0)such that the following integral converges for some values of s.

(1)

The integral (1) depends on the function f and on the number s. Weregard (1) as determining a new function] associated with f, where] isdefined by

(2)

for those values of s for which the integral converges. The function] iscalled the Laplace transform of f. We will also use the notation

.£(j(x) = ](s)

to indicate the relationship (2). Thus.£ is an operator, defined on a classof functions f that we will call object functions, with values] which arefunctions that we will refer to as transforms.

The following three theorems give properties of .£ which are of primaryimportance in the application of Laplace transforms to linear differential

- equations.

THEOREM 1. .£ is a linear operator; i.e., for any s such that](s) and O(s)are both defined, and any numbers a and b,

.£(af(x) + bg'x) = a.£(j(x) + b.£(g(x) = a](s) + bg(s).


Proof. This theorem is just a restatement of the linear properties of thedefining integral (2).

The following theorem will be proved in the next section.

THEOREM 2. If f and g are contin'l,lous on [0, (0) and J(s) == O(s), thenf(x) == g(x).

The point of Theorem 2 is that the operator ..c sets up a one-to-onecorrespondence between continuous object functions f and their transforms1. In other words, if we can determine what J is, then we know what fmust be.

THEOREM 3. If f' is continuous on [0, (0) and limx_ oo e-SXf(x) = 0,then ..c(j'(X)) exists at s if and only if ..c(j(x)) exists at s, and

..c(f'(x)) = s..c(j(x)) - f(O) = sJ(s) - f(O). (3)

Proof. Let s be any number such that limx _ oo e-SXf(x) = O. Integrationby parts gives the following formula for any number R > 0:

Since e-SRf(R) ~ 0 as R ~ 00 by hypothesis, we have

R Rlim r e-sx!'(x) dx = -f(O) + lim s r e-8Xf(x) dx.R_oo Jo R_oo Jo

That is, if either limit above exists, so does the other, and the given equalityholds. This says that ..c(f'(x)) exists at s if and only if ..c(j(x)) does, and ifeither transform exists, then (3) holds.

(4)..c (foX f(t) dt) = ~ ..c(j(x)).

COROLLARY. Assume f is continuous on [0, (0) and F(x) = fof(t) dt.If J(s) exists and limx_ oo e-aXF(x) = 0, then Pes) exists and Pes) =

(1/s)J(s), that is,

Proof. This follows from Theorem 3, since F'(x) = f(x) and F(O) = O.

Note that differentiation of an object function corresponds to an algebraic operation (3) on its transform. We show below how this principle

5-2] THE LAPLACE TRANSFORM 121

is used to solve a linear differential equation with constant coefficients.First we compute the transforms of some simple functions in the followingexamples.

EXAMPLE 1.

£(1) = .!.8

for 8 > O.

Directly from the definition and the assumption that 8 > 0, we have

£(1) = t e-8x1 dx = lim - .!. (e- sR - 1) = .!..} 0 R-+", 8 8

It is clear that the integral above diverges for 8 .:::; 0, so £(1) is definedonly for 8 > O.

EXAMPLE 2.1

£(x) = -8 2 and for 8 > O.

These formulas can be found directly from the definition as in the example above, but we will instead derive them using the corollary ofTheorem 3. Let f(x) = 1, so that fof(t) dt = x = F(x). For every8 > 0, limx-+", e-8XF(x) = limx-+", xe-sx = 0, so

1£(F(x) = - £(j(X)i

8

that is,- 1 1

£(x) = - £(1) = -.8 8 2

Similarly, if f(x) = 2x, then F(x) = x 2, and again

if 8 > o.

Hence

X-+'" X-+'"

1 2£(x2

) = - £(2x) = - £(x)8 8

2 1 2

The transforms of xn , n = 1, 2, 3, ... , can be calculated by repeatingthis process (Problem 2).

EXAMPLE 3.

£(eX) = _1_._

8 - 1 for 8 > 1.

122 TUE LAPLACE TRANSFORM [CHAP. 5

Assume that s > 1, so that e(l-slR ~ 0 as R ~ 00. Then we have

= lim _1_ (e(1-s)R _ 1) = _1_.R-+oo 1 - s s - 1

Now we can illustrate how the Laplace transform is used to solve alinear differential equation. Consider the equation

y' - y = 1 - x, yeO) = 2. (5)

Let y be the solution of (5), and assume* that £(y(x)) exists andlim",-+oo e-S"'y(x) = 0 for all sufficiently large values of s. Then £(y'(x))exists, and

£(y'(x) - y(x)) = £(1 - x),

£(y'(x)) - £(y(x)) = £(1) - £(x),

1 1s£(y(x)) - yeO) -; £(y(x)) = - - 2'

s s

1 1£(y(x))(s - 1) = -; - S2 + 2,

( )s - 1 + 2s 2 1 2

£ y(x) = S2(S - 1) = S2 + S - l'

Since

£(x) = 1/s2, and £(e"') = l/(s - 1),

we have

£(y(x)) = £(x + 2e"').

From Theorem 2 it follows that

y(x) = x + 2e"'.

The use of the transform to solve a linear equation, as illustrated above,can be thought of in the following way. Imagine (Theorem 2) a table inwhich we can find y(x) given yes). By Theorems 1 and 3, a constant co-

* We will show in the next section that these assumptions are always satisfiedfor a solution of a linear equation with constant coefficients.

5-2) THE LAPLACE TRANSFORM 123

efficient linear differential equation in y(x) is transformed into an algebraicequation in yes), involving the given initial condition. We solve thealgebraic equation to find y(s) , consult the table to find y(x), and the problem is solved. Note that the particular solution for the given initialcondition is found without first finding the general solution. If we leavethe initial condition arbitrary, yeO) = c, then we obtain the generalsolution y = x + ce".

PROBLEMS

1. Show by induction that if s > 0, then limx->oo xne-'X = 0 for all positiveintegers n. [Hint: Write xn +1e-·x = xn +1Ie'" and use I'Hospital's rule.)

2. Use Problem 1 and Theorem 3 to prove that £(xn ) = n!jsn+1 for s > 0and n = 0, 1, 2, ....

3. (a) Show that £(eax) = I/(s - a) if s > a.(b) Find £(cosh ax) and £(sinh ax) from part (a) and Theorem 1.

4. Show directly from the definition (2) that

£(sin (ax)) = al(a2 + S2) if s > o.

5. Use Problem 4 and Theorem 3 to find £(cos ax).6. Let f(x) = xeax, so that f'(x) = af(x) + eax, and £(f'(x)) = a£(j(x)) +

I/(s - a). Use this equation and Theorem 3 to find £(xeax) without integration.7. (a) Find £(x2eax ) by the technique of Problem 6.

(b) Show by induction that £(xneax) = n!j(s - a)n for s > a, andn = 1,2, ....

8. Solve the equation y' + y = 1, yeO) = 2, by showing that

1 + 2s 1 1yes) = s(s + 1) = "8 + s + 1 .

[cf. Problem 3(a).]9. Use the transform to solve y' - 2y = 1 - 2x, yeO) 1.

10. Show that with suitable hypotheses,

£(y"(x)) = s2£(y(x)) - sy(O) - y'(O).

11. Find yes) if y is the solution of y" + 2y' + y = eX + 1, yeO) 1,y'(O) = O.

ANSWERS

3. (b) £ (cosh ax) = sl(s2 - a2), £ (sinh ax) = al(s2 - a2)

5. £ (cos ax) = sl(s2 + a2) 6. £(xeax) = 1/(s - a)27. £(x2eax) = 2/(s - a)3 8. y = 1 + e-X

9. y = x + e2x

11. yes) = (s3 + s2 - 1)/s(s - l)(s + 1)2


5-3 Properties of the transform. The discussion of the preceding sectionis largely formal, and in this section we prove the theorems which allowus to deal with transforms in a systematic way.

.We will assume from the beginning that all object functions are continuous on [0, (0). This is not a necessary restriction, but it will simplifythe discussion. A continuous function will have a transform defined forsome values of s provided the function does not grow too rapidly as x --+ 00.

We say that f is of exponential order b if e-bXf(x) is bounded on [0, (0);i.e., if there is a number B such that e-bXlf(x)! S B for all x ~ 0.

THEOREM 1. If f ~'s of exponential order b, then !(s) exists for all s > b.

Proof. If s > b, then s = b + p for some p > 0, and

Since p > 0, the integral

~oo Be-pxdx

converges, and therefore the integral. for !(s) converges by Theorem 1,Section 5-1.

We restate Theorem 3 of the last section and its corollary here in termsof our standard hypotheses of continuity and exponential order.

THEOREM 2. If f and f' are continuous and f is of exponential order b,then £(j'(x) exists for s > b, and

£(j'(x) = s£(j(x) - f(O).

Proof. If f is of exponential order b, then limx-->oo e-SXf(x) = °for s > b(Problem 1), and this is the condition of Theorem 3, Section 5-2.

COROLLARY. If f is of exponent~'al order b and

F(x) = Jof(t) dt,

then F ~'s of exponential order band

£(F(x) = (I/s)£(j(x).

Proof. By the corollary of Theorem 3, Section 5-2, it is sufficient to showthat F is of exponential order b. Write

5-3]

and conclude that

PROPERTIES OF THE TRANSFORM 125

IF(x) I s i X

Bebt

dt = ~ (ebx - 1).

Therefore, e-bXIF(x)! S B/lbl, and F is also of exponential order b.

If the functions f, f', f" etc., are all continuous and of exponential orderb, then by repeated applications of Theorem 2, we obtain the followingbasic formulas for s > b:

.£(f'(x) = sJ(s) - f(O),

.£(j"(X) = S2J(S) - sf(O) - f'(0), (1)

.£(j/"(x) = s3j(s) - s2f(0) - sf'(O) - f"(0), etc.

THEOREM 3. If f 1'S of exponential order b, then e-aXf(x) is of exponential order b - a, and for s > b - a, we have

Proof. Problem 2.

.£(e-aXf(x) = J(s + a). (2)

(s > 0).

As an example of (2), consider the formula (see Problem 4, Section 5-2)

.£ (sin bx) = S2 ~ b2 '

From this and (2) we have immediately

( -ax . b) b.£ e sm x = (s + a)2 + b2 '

Similarly, from1

.£(x) = 2"' (s > 0),s

(s > -a) .

we get (with a = -1)

.£(xeX) = (s ~ 1)2' (s > 1).

Formulas (1) describe what happens to the transform when the objectfunction is differentiated. The next theorem describes what happens tothe object function when the transform is differentiated. More precisely,we show that each transform J is differentiable, that its derivative J/ isagain a transform, and that

J/(s) = .£(-xf(x).

The proof of these facts depends on the following lemma.

(3)


LEMMA 1. If f is of exponential order b, and n is a positive integer, then£(xnf(x» exists for s > b, and the ~'ntegral for this transform convergesuniformly for s ~ bo, for any bo > b.

Proof. Problem 3.

Notice that the lemma above implies that the integral for £(xnf(x»converges uniformly on some interval around s, for each s > b.

THEOREM 4. If f 1'S of exponential order b, then 1 has derivatives of allorders for s > b, and

1'(s) = £( -xf(x» ,

1"(s) = £(x2f(x» ,

1 'I1 (s) = £(-x3f(x», etc.

(4)

Proof. The formulas (4) are obtained by repeatedly differentiating theintegral formula for l(s) under the integral sign. For example,

l(s) = 1000

e-OXf(x) dx,

1 ' (s) = 1000

e-OX (-x)f(x) dx.

(5)

(6)

(s > a).

Differentiation under the integral sign in (5) is justified by Theorem 4,Section 5-1, since by the lemma we know that the resulting integral (6)converges uniformly on some interval around each s > b.

As an example of how new transforms can be obtained from (4), considerthe formula (see Problem 3, Section 5-1)

£(eaX) = _1_,

s-a

By differentiating the right side, we get

( ax) 1.£ xe = (s _ a)2 '

and in general

n( n-l ax) (n - I)!"'" X e = ,(s .,- a)n

In a similar way, we can start with

(s > a),

(s > a).

£(1)1= -,S

(s > 0)

5-3] PROPERTIES OF THE TRANSFORM 127

and obtain the formulasn!=-,Sn

(s > 0).

Since our purpose is to solve linear differential equations with constantcoefficients by means of transforms, we need to know that the solutionswe seek always have transforms. This question is settled by the followinglemmas and theorem.

LEMMA 2. If fI, ... ,fn are of exponential orders bl , ... , bn, respectt'vely,then fx + ... + f n is of exponential order b = max IbI, ... , bn I.

Proof. Problem 4.

LEMMA 3. If q, g', q", ... , q(k) are continuous and of exponential orderb, and y is a solution of

(D - r)y = q, (7)

then y, y', ... , y(k+l) are continuous and of exponential order b' =

max Ib, rI·

Proof. Any solution y of (7) can be written

y(x) = cerx + erxlX e-rtq(t) dt.

The solution y will have the required properties if the integral

u(x) = { e-rtq(t) dt

(8)

(9)

has k + 1 continuous derivatives which are of exponential order b - r.The function u itself is of order b - r, by the corollary of Theorem 2,since the integrand in (9) is of order b - r. Also, u is differentiable (hencecontinuous), and its derivatives are given by

u'(x) = e.-rXq(x) ,

u"(x) = e-rX[q'(x) - rq(x)),

u"'(x) = e-rX[q"(x) - 2rq'(x) + r2q(x)), etc.

(10)

From (10) it is clear that u, and hence y, has one more continuous derivative than q. From Lemma 2 and (10) we see that u', u", ... , U(k+ll areof exponential order b - r, since the terms in brackets are of or{iler b.

If r is a complex number, we modify the proof above by c'onsideringthe real part of r, and the lemma holds for real or complex numbers r.


THEOREM 5. If q is continuous and of exponent1'al order, and Y is asolution of

y(n) + Pn_Iy(n-ll + ... + PlY' + PoY = q, (11)

where Po, PI, ... , Pn-l are constants, then Y, y', ... , y(n) are continuous,and of exponential order.

Proof. We write Eq. (11) in the form

(D - rl)(D - rz) ... (D - rnlY = q (12)

and assum,e that Y is a given solution of (12). Define functions YI, Yz, ... ,Yn-l by

(D - rz)(D - ra) (D - rn)y = YI,

(D - ra) (D - rn)y = Yz,

(D - rn)y = Yn-l.

From these equations, we have

(D - rl)YI = q,

(D - rZ)Yz = YI,

(13)

(D - rn-I)Yn-1 = Yn-Z,

(D - rn)y = Yn-l.

By Lemma 3 and the first of equations (13), YI and y{ are continuous andof exponential order. From the second equation it then follows thatYz, y~, and y~' are continuous and of exponential order. Continuing inthis way, we see that Yn-I, Y~-I, ... , Y~":.:il) are continuous and of exponentialorder. The last equation in (13) and Lemma 3 then guarantee thatY, y', ... ,y(n) are continuous and of exponential order.

COROLLARY. If q is cont1'nuous and of exponential order, and Y is asolution of (11), then o£(y(x», o£(y'(x», ... , o£(y(n)(x» exist for allsufficiently large values of s.

The fact that £ is a one-to-one operator is fundamental in our applications. Using the following theorem, we will prove this next.

THEOREM 6 (Weierstrass polynomial approximation theorem). If u iscont£nuous on [0, 1] and E > OJ then there is a polynomial P such thatlu(x) - P(x)1 < E for all x in [0, 1].

5-3] PROPERTIES OF THE TRANSFORM 129

Theorem 6 is one of the most useful theorems of analysis. Althoughthe statement of the theorem is quite simple, the proof is complicated andwill be omitted.

COROLLARY. If u is continuous on [0, 1] and I5 xnu(x) dx = 0 forn = 0, 1,2, ... , then u(x) == o.

Proof. Let E be any positive number, and let P be a polynomial suchthat lu(x) - P(x)1 < E for all x in [0, 1). By the hypothesis,

101

P(x)u(x) dx = 0,

and hence

{I (I 11

0u(x) 2 dx = 1

0u(x)[u(x) - P(x)] dx :s; E10 lu(x)1 dx.

Since E is arbitrary, the integral of u(x) 2 is zero, and hence u(x) == O.

THEOREM 7 (.c is one-to-one). Iff and g are continuous on [0, 00) and!(s) = O(s) for all s ?: so, then f(x) == g(x).

Proof. It is sufficient (Problem 6) to show that if h is any continuousfunction such that h(s) = 0 for s ?: so, then hex) == O. Assume thereforethat h(s) = 0 for s ?: so, and in particular that h(so + n) = 0 forn = 0,1,2, ... :

(14)

Define the function v as follows:

(15)

Note that v is continuous on [0, 00), with v(O) = 0 and lim"'....."" vex) =h(so) = O. Now integrate (14) by parts, with u = e-n", and dv =e-So"'h(x) dx, so that v is given by (15). We get

o = h(so + n) = lim [e-nRV(R) + n rRe-nxv(x) dX]

R....."" 10

= n10"" e-nxv(x) dx;

that is, for n = 0, 1, 2, ... , we have

(16)


We make a change of variable in (16), with e-X = t, e-nx = tn, X = In t-I,dx = -(lit) dt, and let u(t) = v(ln t-I) = vex). As x ranges between °and 00, t ranges between 1 and 0. If we let u(o) = 0, then u is continuouson [0, 1], since vex) ~ °as x ~ 00 (t ~ 0). With this change of variable,(16) becomes

fo 1tn-Iu(t) dt = 0.

Therefore u(t) == °on [0,1), and vex) == °on [0, (0). Hence v/(x) =e-SoXh(x) == 0, and hex) == 0.

Since £ is one-to-one, we can define the inverse operator £-1 whichmaps transforms onto object functions. In other words,

£-l(4)(S)) = f(x)

if and only if £(J(x)) = 4>(s). It is easy to show (Problem ]2) that £-1is also a linear operator; i.e., that

In some applications it is convenient to allow functions q in (11) whichhave discontinuities. For example, such a function would arise in thediscussion of an electric circuit in which a constant voltage was appliedat some time to > 0. Although the theorems of this section are statedand proved for continuous functions, they extend with only minor changesto functions with jump discontinuities at isolated points.

PROBLEMS

1. Show that if f is of exponential order b, then limx-+oo e-Sxf(x) = 0 for alls > b.

2. Prove Theorem 30*3. Prove Lemma 1. [Hint: Assume that bo > b, and let p = bo - b > 00

Show that if s ~ bo, then e-Sxxnlf(x) I :s:; Be-pxxn. Then use Theorem 3, Section5-1, and Problem 2, Section 5-2.]

4. Prove Lemma 2050 Explain the difference between the Weierstrass polynomial approximation

theorem and the false statement that every continuous function has a powerseries expansion.

6. (See Theorem 7.) Show that f(s) == O(s) implies f(x) == g(x) if and onlyif h(s) == 0 implies hex) == o.

70 Show that £(e-X2 ) exists for every value of s. [Hint: Show that e-axe-x2 :s:;e-X if x ~ lsi + 1, and use Theorem 1, Section 5-1.]

8. Show that £(eX2 ) does not exist for any value of s. [Hint: Show that forevery s, limx-+oo e-sxex2 = 00 0]

5-4] SOLUTION OF EQUATIONS BY TRANSFORMS 131

9. For any number a > 0, let fa be defined as follows:

fa (x) = (f(X : a) if x 2': a,

if 0 ~ x < a.

Show that £ (ja(X)) = 1(s + a).10. Let f be the function which is zero on [0, 1) and one on [1, 00). Find

£(j(x)).11. All the formulas of Table 2, Section 5-4, can be derived from Formulas

1 and 9 and the theorems of this section. Without using the definition, derive

.(a) Formulas 2 through 4.(c) Formulas 10 through 12.

(b) Formulas 5 through 8.(d) Formulas 13 through 15.

12. Show that £-1 is a linear operator; i.e., verify formula (17).

5-4 Solution of equations by transforms. Table 1 is a list of the operational formulas which were proved in Section 5-3 under appropriateassumptions, and one formula, (E), which is discussed below. Table 2 is ashort table of Laplace transforms of specific elementary functions. Theobject functions listed in Table 2 are all of exponential order, and henceeach of the transforms is defined for all sufficiently large values of s. Westart with some examples to show how such a table is used to obtain solutions of linear equations with constant coefficients and specified initialconditions.

EXAMPLE 1. y" + 4y = x + sin 2x, y(O) = y' (0) = O.

Let y be the solution of this equation, and y be its transform. For simplicity, we will write y instead of y(s) in the transformed equation inthe same way that y, y" are used for y(x), y"(x) in the differential equation

TABLE 1

A £(!'(x)) = sl(s) - f(O)

B £ (lX f(t) dt) = ~ l(s)

c £(e-axf(x)) = l(s + (I)

D £(-xf(x)) = 1 ' (s)

E £(f(xhg(x)) = j(s)g(s)

132 THE LAPLACE TRANSFORM

TABLE 2

[CHAP. 5

l(s) f(x)

11

1-s

1 1 n-l (n = 1,2,3, ...)2 - xsn (n - I)!

1 ax3 -- e

s-a

1 1 n-l ax(n = 1,2,3, ...)4 X e

(s - a)n (n - I)!

51 1 . h

S2 - a2- sm axa

6s cosh ax

S2 - a2

1(a ,e b) 1 [ ax bXj

7 -- e -e(s - a)(s - b) a-b

8s

(a ,e b) 1 [ax b bXj-- ae - e(s - a)(s - b) a-b

1 19

S2 + a2 - sin· axa

s10

s2 + a2cos ax

1 1 ax sin bx11(s - a)2 + b2 be

s - a ax cos bx12(s - a)2 + b2 e

s 113 (s2 + a2)2

2a x sin ax

2 2S -a

14(s2 + a2)2

x cos ax

1 2~3 [sin ax - ax cos ax]15 (s2 + a2)2


itself. The transform '0 must satisfy

82'0 - 8Y(0) - y' (0) + 4'0 = 812 + 82 ~ 4 '

or, since yeO) = y'(O) = 0,

'0(82 + 4) = 8\ + 82~ 4 '

• 1 + 2y = 82(82 + 4) (82 + 4)2

Hence

y = £-1 t2(8/+ 4)} + 2£-1 {(82 ~ 4)2} .

From Table 2 (#15), we find

2£-1 {(82~ 4) 2} = t [sin 2x - 2x cos 2x]. (1)

Since 1/82(82 + 4) does not appear in the table, we break this expressioninto partial fractions:

1 A B C8 + D82(82 + 4) = -i + 82 + 82 + 4

Equation (2) is an identity if

1 == (A + C)83 + (B + D)82 + 4A8 + 4B.

Hence B = 1, A = 0, D = -1, C = 0, and

1 1 1 1 182(82 + 4) = 4: 82 - 4 82 + 4 .

From the table (#2, #9), we get

-1 { 1 } _ 1. 1 1 . 2£ 8 2(82 + 4) - 4"X - 4" 2" sm x.

Finally, from (1) and (3), we have

y = t [sin 2x - 2x cos 2x] + 1x - t sin 2x

= 1x - 1x cos 2x.

(2)

(3)


EXAMPLE 2. y" + 2y' + 2y = 2e-x cos x, y(O) = 2, y'(O) = -2.

Taking the transform of both sides, using #12, we get

Collecting terms and simplifying, we obtain

A 2 2(8 + 1)Y(8 + 28 + 2) = (8 + 1)2 + 1 + 2(8 + 1),

A 2(8 + 1) 2(8 + 1)y = [(8 + 1)2 + 1]2 + (8 + 1)2 + 1

Note that the first term on the right of (4) can be written

(4)

2(8 + 1)[(8 + 1)2 + 1]2

d -1d8 (8 + 1)2 + 1

Therefore, using (D), then #11 and #12, we have

-1 { -1 } + 2£-1 { 8 + 1 }y = -x£ (8 + 1)2 + 1 (8 + 1)2 + 1

= xe-x sin x + 2e-x cos x.

Instead of using (D) as above, we can use (C), and then #13, to obtain

-1 { 2(8 + 1) } 2 -x£-1 { 8 }£ [(8 + 1)2 + 1]2 = e [82 + 1]2

Because of the form of the entries in Table 2, it is usually necessary tobreak a transform into partial fractions before we can identify the inversetransform. Next we give a simple method for effecting the partial fractiondecomposition for a rational function P(8)jQ(8) when Q(8) is a product ofdistinct linear factors. Suppose that Tl, ... , Tn are distinct numbers, that

Q(8) = (8 - Tl) ... (8 - Tn),

and that P is a polynomial of degree less than n. Then there are numbersA 11 ••. , An such that

P(8) =~ + ... +~.Q(8) 8 - Tl 8 - Tn

(5)


If we multiply both sides of (5) by s - Ti, and let s approach Ti, only Airemains on the right, and hence

Ai = lim (s - Ti)P(S) .S->Ti Q(s)

Since Q(Ti) = 0, we can write (6) in the form

Therefore the decomposition (5) can be written

EXAMPLE 3. Express in partial fractions

S2 - 2s + 2

(6)

S3 - S2 - 4s + 4

Here Q(s) = S3 - S2 - 4s + 4 = (s - l)(s - 2)(s + 2), and Q/(s) =

3s2 - 2s - 4. Hence P(l)/Q'(l) = 1/(-3) = -t, P(2)/Q'(2) = t =!, P(-2)/Q'(-2) = ~g = t, and

S2 - 2s + 2 = _ .!. _1_ + .!. _1_ + ~ _1_ .S3 - S2 - 4s + 4 3 8 - 1 2 8 - 2 6 8 + 2

EXAMPLE 4. y" - 2y' - 3y = eX, yeO) = 1, y'(O) = 1.

Q(8) = 83 - 382- 8 + 3, and Q/(8) =

P(l)/Q'(l) = -i, P(3)/Q'(3) = i,

Taking the transform of both sides and simplifying, we get

82y - 8 - 1 - 2(8Y - 1) - 3y = _1_ ,8 - 1

A( 2 2 3) 1 + 1 82

- 28 + 2ys - s- =8-1 s- = 8-1 '

A S2 - 28 + 2Y = (8 - 1) (8 - 3) (8 + 1) .

Here P(8) = 82- 28 + 2,

382- 68 - 1. Hence

P(-l)/Q'(-l) = ~, and

A = 82

- 2s + 2 = _ .!. _1_ + ~ _1_ + ~ _1_.Y (8-1)(8-3)(8+1) 48-188-388+1

136

Therefore

THE LAPLACE TRANSFORM [CHAP. 5

We continually use the linear properties of "c-that is, we continuallyuse the fact that "c is a mapping which preserves the operations of additionof functions and multiplication of a function by a constant. There is a.type of multiplication for object functions, denoted f*g, for which "c is alsoa multiplicative mapping, in the sense that

"c[(f*g)(x») = "c(j(x»)"c(g(x»).

The product on the right above is the usual pointwise product of twofunctions. The "product" f*g, called the convolution of f and g, is defined by

(7)

For convenience in discussing the convolution of particular formulas, wewill usually use the slightly improper Dotationf(x)*g(x) instead of (f*g) (x).

We give here without proof a formal statement of the multiplicativeproperty of "c with respect to convolution.

THEOREM 1. If f and g are conMnuous on [0, 00) and of exponenMal orderb,' then f*g is continuous and of exponent1"al order b, and for all s > b

"c[f(x)*g(x»). "c(j(x) )"c(g(x»). (8)

Convolution has the following properties, which justify considering thisoperation as a type of product for functions.

f*g = g*f,

(f*g)*h = f*(g*h),

f*(g + h) = f*g + f*h,

(cj)*g . f*(cg) = c(f*g).

(9)

The formulas above can be derived directly from the definition (7), butfollow most easily from (8) and the fact that "c is one-to-one. For example,from (8) we have

"c[f(x)*g(x») = "c(j(x) )"c (g (x) )

and

"c[g(x)*f(x») = "c(g(x»)"c(j(x»).


The right sides of the two equations above are obviously equal, so

Since £ is a one-to-one operator, it follows that

We list next some examples of the convolution of two functions and theparallel statements for their transforms.

14(x) = fo'" f(t) dt,

£(hf(x)) = £(I)£(J(x)) = .! £(J(x)).8

x* sin x = fo'" (x - t) sin t dt = x - sin x,

£(x* sin x) = £(x)£ (sin x) = 812 8 2 ~ 1 = 812 - 82 ~ 1 .

x*e'" = 10'" (x - t)e t dt = e'" - x-I,

£(x*e"') = £(x)£(e"') = 1-. _1_ = _1__ 1- _ .!.8 2 8 - 1 8 - 1 8 2 8

sin x* cos x = fo'" sin (x - t) cos t dt = !x sin x,

,£ (sinx* cos x) = £ (sinx)£ (cos x) = 82 ~ 1 . 82 ~ 1 = (82 ~ 1)2

(10)

(11)

(12)

(13)

(14)

EXAMPLE 5. y" + y = cos x, y(O) = y' (0) = O.

The transformed equation is

2. + . 88 Y Y = 82 + 1 '

or

y = 8 2 ~ 1 . 82 ~ 1 = £ (sin x)£ (cos x).

It follows from (8) and (14) that

y = sin x* cos x = !x sin x.


EXAMPLE 6. y" - 5y' + 6y = e-2x,

The transformed equation is

yeO) = 0, y/(O) = -2.

YA(82 _ 58 + 6) = _1_ - 28 + 2 '

or

Y = (8 + 2)(8 -= 2)(8 - 3) - (8 - 2)2(8 - 3) . (1-5)

From Table 2 (#7), we get

-1 { 1 } 3x 2x£ (8 - 2)(8 - 3) = e - e .

Therefore

-1 { 1 1} -2x [ 3x 2Xj£ 8 + 2 . (8 - 2) (8 _ 3) = e * e - e

= lX e-2(x-t)[e3t - e2t j dt

= e-2xlX e2t[e3t - e2tj dt

= e-2X[ie 5X - t - i-e4X + il_ .l.e3x 1 2x l -2x- 5 - 4 e - 20e .

Combining the results of (15), (16), and (17), we have

_ £-1 { 1.}_2£-1 { 1 }y- (8+2)(8-2)(8-3) (8-2)(8-3)

= te3X _ i-e 2x - -i'oe-2x _ 2[e3X _ e2X j

-te3X + ie2x - -i'oe-2x.

PROBLEMS

Solve the following equations using transforms.

1. y/ - Y = eX, yeO) = 12. y/ = X, yeO) = 13. y" - 3y' + 2y = e-x , yeO) = -1, y/(O) = 14. y" + 9y = sin 3x, yeO) = 1, y/(O) = 05. y" + 2y' + y = 1 + eX, yeO) = 1, y/(O) = 06. y" + y = x + eX, yeO) = 2, y/ (0) = 17. y" - 3y' + 2y = 2e3x, yeO) = 2, y/(O) = 3

(16)

(17)


8. y'" - y' = e2x, yeO) = y' (0) = y" (0) = 09. y" + y' = 3x2 - 6, yeO) = 0, y'(O) = 1

10. y'" - y' = 2 sin x, yeO) = y' (0) = y" (0) = 011. y" + 2y' + 2y = 0, yeO) = y'(O) = 112. y" - 2y' + y = eX sin x, yeO) = y'(O) = 013. y" - 4y' + 5y = e2x cos x, yeO) = y'(O) = 0

14. Verify the convolution formulas in (10), (11), (12), (13), and (14). Checkin each case that £[!(x)*g(x)] = £(j(x»)£(g(x»).

15. Make the change of variable t = x - tt, dt = -du in (7) and verifythat !*g = g*f.

16. (a) Show that!(ax)*g(ax) = (l/a)(f*g)(ax).(b) Use (a) and (14) to conclude that (sin 2x)* (cos 2x) = !x sin 2x.(c) Use (b) to solve y" + 4y = cos 2x, yeO) = y' (0) = O.

17. (a) Use the definition (7) to show that

(sin axh (sin ax) = 2~ sin ax - !x cos ax.

(b) Derive the formula of (a) using transforms (#9 and #15 of Table 2).18. (a) Use the definition (7) to show that

(cos axh (cos ax) = !x cos ax + 21a sin ax.

(b) Derive the formula of (a) using transforms (#10, #14, and #15of Table 2).

ANSWERS

1. y = (1 + x)eX

3. y = ie2x - ~ex + ie-x

5. y = 1 + i-ex - i-e-x - !xe-x

7. y = 2ex - e2x + e3x

9. y = 1 - 3x2 + x3 - e-X

11. y = e-Xcos x + 2e-Xsin x13. y = !xe2x sin x

2. Y = 1 + !x2

4. Y = ls sin 3x+ cos 3x - ix cos 3x6. y = x + !ex+ ~ cos x - ! sin x8. y = ie2x - !ex - ie-x + !

10. Y = 2 - sin i + 2e - e-X

12. y = xeX - eX sin x16. (c) y = i-x sin 2x

\

CHAPTER 6

PICARD'S EXISTENCE THEOREM

6-1 Review. We will start this chapter with a review of some of thebasic definitions and theorems from calculus which will be used in theexistence proof.

A function f of one variable is continuous at a if for every E > 0 there isa 0 > 0 such that If(x) - f(a) I < E whenever Ix - al < o. The function f is continuous on an interval I if f is continuous at each point of I. Afunction F of two variables is continuous at (a, b) if for every E > 0 thereis a 0 > 0 such that IF(x, y) - F(a, b)1 < E whenever Ix - al < 0 andIy - bl < o. The function F is continuous on a set of points (e.g., a square)S if F is continuous at each point of S.

THEOREM 1. A function of one variable which is continuous on (L closedinterval is bounded on that interval, and sim1'larly a function of two variableswhich is continuous on a closed square (inCluding the boundary lines) 1'Sbounded.

We will use the notation

max If(t)!tin [a,b]

for the least upper bound of the numbers If(t) I for t in [a, b].·The student is familiar with the basic properties of the definite integral

lab f(t) dt. (1)

We recall that there are many functions f for which (1) is not defined, butthat (1) is defined for any function f which is continuous on [a, b] (or[b, a], if b < a). The following is essentially the fundamental theorem ofcalculus.

THEOREM 2. Iff is continuous on [a, b] and g is defined on [a, b] by

g(x) = {; f(t) dt,

then g 1'S differentiable, and g'(x) = f(x) for x in [a, b].

We will also use the following two integral inequalities.140

6-1] REVIEW 141

THEOREM 3. IJ J is continuous on [a, b] (or [b, a]), then

ItJ(t) dtl :::; It IJ(t)1 dtl

and

I(b J(t) dtl :::; Ib - aJ max IJ(t) I·Ja tin [a,b]

(2)

(3)

The Picard method "constructs" a solution to the differential equationin question as the limit of a sequence of functions. We turn to some factsabout sequences.

The sequence (of numbers) Ian} converges to a (denoted limn-->oo an = a)if for every positive number EO there is an integer N such that Ian - al < EO

if n ~ N.If IYn} is a sequence of functions defined on a given interval I, then

we can regard this as a family of sequences of numbers; i.e., for each fixedx in I, we hav~ the sequence IYn(x)} of numbers. The limit of {Yn(x)}, if itexists, will of course depend on x, and hence a function Y is defined byy(x) = limn-->oo Yn(x). If the sequence of numbers IYn(x)} converges foreach x in I, we say that IYn} converges to Y pm'ntwise on I and denote thislimn-->oo Yn(x) = y(x) (x in I). Simply writing out the formal e-definitiongives us the equivalent statement: {Yn} converges to Y pointwise on I if forevery x in I and every positive number EO, there is an integer N (dependingon both x and EO) such that IYn(x) - y(x)1 < EO if n ~ N.

Now we forget temporarily the preceding definition and ask what kindof geometric interpretation should go with the idea of convergence of a.sequence of functions. One reasonable idea is that the graphs of the functions Yn should approach the graph of the limit function y. This is a sensibleidea, but it is not a consequence of the definition of pointwise convergence.

G'l)(to>!) I G.l) (1, 1)

(to, 0) a,o) (1,0) (2,0)

FIG. 6-1. Functions converging pointwise but not uniformly.

142 PICARD'S EXISTENCE THEOREM [CHAP. 6

EXAMPLE 1. (d. Fig. 6-1.) Let Yn (n= 1,2, ...) be the function on[0,2] whose graph consists of the line segments from (0,0) to (lin, 1),from (lin, 1) to (2In, 0), and from (2In,0) to (2,0). This sequence converges to the function identically zero on [0,2] (Problem 2), even thougheach function is at some point one unit away from the x-axis.

We define next a type of convergence for sequences of functions whichis a formalization of the statement that the graphs of the individualfunctions approach the graph of the limit function.

The sequence IYn l of functions on I converges to y uniformly on I if forevery positive number E, there is an integer N (which depends on E but noton x) such that IYn(x) - y(x) I < E for all x in I, if n ~ N.

It is clear that uniform convergence is a stronger hypothesis thanpointwise convergence. Some of the convenient properties of uniformconvergence-and significant reasons for making such a definition-aregiven in the next two theorems.

THEOREM 4. If IYn l is a sequence of continuous functions on [a, b] whichconverges uniformly on [a, b] to Y, then Y is cont1'nuous on [a, b].

This theorem is not true for pointwise convergence (Problem 4).

THEOREM 5. If IYnl is a sequence of continuous functions on [a, b] whichconverges uniformly on [a, b] to Y, then

b lb!~ fa Yn(t) dt = a y(t) dt.

This theorem also is false with the weaker hypothesis of pointwiseconvergence (Problem 3). The student is asked to supply a proof ofTheorem 5 in Problem 6.

Now recall the connnection between series (infinite sums) and sequences.We say the series I::=l ak converges to a (denoted I::=l ak = a) if the

sequence ISn l converges to a, where

Sn = al + a2 + ... + an'

The sequence ISnl is called the sequence of partial sums of 2::=1 ak·If IUk l is a sequence of functions defined on some interval I, we can

form the series with these terms, I::=l Uk. The convergence of a seriesof functions depends in the obvious way on the convergence of the sequenceISnl of functions which are the partial sums: Sn(x) = Ul(X) + ... + un(x)for x in I. Thus we agree that I::=l Uk converges pointwise on I if thesequence ISnl converges pointwise on I, and I::=l Uk converges uniformlyon I if ISnl converges uniformly on I. We will use the following comparison test.

6-1] REVIEW 143

THEOREM 6. If for each k, Uk is a non-negat?'ve function on I, and L:=l Ukconverges uniformly on I, and IYk(X) I ::; Uk(X) for each k and each x in I,then L:=l Yk converges uniformly on I.

PROBLEMS

1. Show (given E > 0, find 0 > 0) that if J is continuous at a, and J(a) = b,and F is continuous at (a, b), and get) = F(t,JCt), then g is continuous at a.

2. (a) Show that the sequence of functions {Ynl given in Example 1 convergespointwise to the function identically zero on [0, 2]; that is, give an integer N", for each x in [0, 2] such that IYn(x) - 01 < E if n ~ N ",. [Hint:it is sufficient to give an N", such that Yn(X) = 0 if n ~ N",.]

(b) Show that if the functions of Example 1 are changed in any way onthe intervals on which they are positive (but only there), the newsequence still converges to zero.

3. Modify the functions of Example 1 to obtain a sequence {Yn I of functionssuch that limn->oo Yn(X) = 0 (x in [0, 2]), and nYn(X) dx = 1 for all n [cf.Problem 2(b)].

4. Let Yn be the sequence of functions on [-1,1] defined by

0 if -1 ~x ~ 0,

Yn(X) if 0 ~x1

nx ~ -,n

1 if1-~x~ 1.n

Graph YI, Y2, Y3, and Y4. Show that {Ynl converges on [-1, 1]. Graph the limitfunction. Does the sequence converge uniformly on [-I, I]?

5. Give an example to show that the outside absolute value signs on the rightside of (2) are necessary. [Hint: When is J~ IJ(t)1 dt negative?]

6. Show that if {Ynl is a sequence of continuous functions on [a, b] whichconverges uniformly on [a, b] to y, then

l~ t Yn(t) dt= t Yet) dt.

[Hint: Use (3) with J replaced by Yn - y. Compare this result with Problem 3.J7. Let {8nl be any sequence. Find aI, a2, ... such that {8nl is the sequence

of partial sums for the series L,::=l ak.8. Show that if Uk is a continuous function on [ for each k, and L,::=l Uk

_ converges uniformly on [ to the function u, then U is continuous on [.9. Show that if IYk(X)/ ::; /xJk/k! for all x in [-I, IJ and all k, then L,::=l Yk

converges uniformly on [-I, IJ.10. Let Yn(X) = xn, Zn(X) = xn/n! Discuss the intervals on which the

sequences IYn I and {zn I converge and the intervals on which the sequencesconverge uniformly.

144 PICARD'S EXISTENCE THEOREM [CHAP. ()

6-2 Outline of the Picard method. We will prove in this section andthe next the existence and uniqueness. of a solution of the equation

y'(x) = F(x, y(x»), y(a) = b. (1)

The equation (1) can be written in an equivalent integral form, as we showin Theorem 1, and it is this integral form which is principally used in therest of the chapter.

THEOREM 1. If F is continuous on the square S = {(x, y): Ix - al :::; handly - bl :::; h} andyisacontinuousfunctiononI = {x:lx - al:::; h}whose graph.is contained in S, then y is a solution of (1) on I if and only ify satisfies the following equation ident~'cally on I:

y(x) . b + f; F(t, y(t») dt. (2)

Proof. Notice that the integral on the right side of (2) exists for all xin I, since the integrand is continuous on I (see Problem 1, Section 6-1).

First assume that y satisfies (2) identically on I. Then in particular

y(a) = b + faa F(t, y(t») dt = b, (3)

so y satisfies the initial condition of (1). From the continuity of the integrand in (2) we can conclude that y is differentiable on I (Theorem 2,Section 6-1) and that

y'(x) = F(x, y(x») (4)

for each x in I. That is, (4) is an identity on I, and y satisfies (1) on I.Now suppose y satisfies (1) on I. The integrals of equal functions are

obviously equal, so we have

f y'(t) dt = fax F(t, y(t») dt (5)

for all x in I. The left side of (5) can be integrated to obtain y(x) - y(a) =y(x) - b, so y satisfies (2) on I.

We can now outline the basic i«iea of the Picard method. Define asequence of functions IYn} as follows:

Yo(x) . b,

Yl(X) = b +f F(t, b) dt,

Y2(X) = b + fax F(t, Yl (t») dt,

etc.

(6)

6-2] OUTLINE OF THE PICARD METHOD 145

Suppose there is a limit function Y such that

lim Yn+l(X) = y(x)n->oo

and

(7)

(9)

(10)

Since the left sides of (7) and (8) are equal by definition (6), it follows thatthe limit function Y satisfies

y(x) = b +f; F(t, y(t)) dt

and hence is a solution of (2) and (1).The following three questions must be resolved for this approach to

work:

I. Does the scheme (6) actually define all the functions Yn on somefixed interval 10 around a?

II. Does the sequence defined by (6) converge on I o?III. Does the sequence of functions f~ F(t, Yn(t)) dt converge on 10 as

required by (8)?

Our proof of the existence of a solution will consist in verifying conditionsI, II, and III under appropriate assumptions on F.

EXAMPLE 1. Consider the equation y'(x) = 2xy(x) , y(O) = 1: Theequivalent integral form is

y(x) = 1 + foX 2ty(t) dt.

We let yo(x) = 1, Yl(X) = 1 + f o2tdt, and in general

Yn+l(X) = 1 + foX 2tYn(t) dt.

Here the function F(x, y) = 2xy is continuous everywhere, so all theintegrals in (9) exist for all x, and all the functions Yn are defined everywhere (Condition I). We get the following formulas:

Yl(X) = 1 + fo'" 2t1 dt = 1 + x Z,

yz(x) = 1 + foX 2t(1 + tZ) dt = 1 + XZ + tx4,

y (x) = 1 + XZ + !x4 + _1_ x 6 + ... +..!.. XZnn 2 3·2 n!

~ 1 Z k= LJ k! (x ) .k=O


The functions Yn form the sequence of partial sums of the series for ex2•

This series converges everywhere, which means the sequence {Yn(x) I converges to ex2 for each x. Condition II is therefore satisfied on any intervalaround zero.

Condition III requires that

fx Irx t2 2lim 2tYn(t) dt = 2te dt = eX - 1n--->oo 0 0

for all x in some interval around zero. By (10)

('" 2tYn(t) dt = ('" 2t :t ;, t2k

Jo Jo k=O'

= :t 2, ('" t2k+1 dtk=O k. Jo

f-- 2 1 2k+2

= 6 k! 2(k + 1) x

_ f-- 1 ( 2)k+l- L.J (k + I)! x

k=Q

(11)

(12)

y(x) + 2f yet) dt = 1.

~1= 6 k! (x

2)k.

The functions (12) are the partial sums of the series for ex2- 1, and

therefore converge to this function on every interval as required in (11).Condition III is satisfied and the limit ex2 of the sequence {Yn(x) I is asolution.

PROBLEMS

1. Write a first order differential equation with initial condition which isequivalent to

2. Write a second order differential equation with initial conditions which isequivalent to

[X 2y(x) + x

10Yet) dt = x + 1.

3. Find the differentiable function y which satisfies

[X 2(a) 10 Yet) dt = x - y(x), r'" 2(b) xy(x) +11 yet) dt = x .

6-3] PROOF OF EXISTENCE AND UNIQUENESS 147

4. Compute the sequence {Ynl defined by (6) for the equation y/ - Y = 0,yeO) = 1. Verify that conditions I, II, and III are satisfied for this sequence.

5. Compute the functions Yn defined by

{

YO(X) = 0,

Yn+l (x) = 1 + fox F(t, Yn(t)) dt,

where F(x, y) = y. Compare with Problem 4.6. Find the sequence {Ynl defined by (6) for the equation y/ = Y - x,

yeO) = 1. Verify I, II, and III for any finite interval [-r, r].7: Use the Picard method, as in Problem 6, to solve y/ = x2 - 1 - Y,

.y(0) = 1.8. Show that III is not an automatic consequence of II. [Hint: Let F(x, y) = Y

and use the sequence {Yn} of Problem 3, Section 6-1.]9. Let {Yn} be a sequence of continuous functions on [0, 1] which converges

uniformly on [0, 1] to the function y. Let F(x, y)= xv. Show that for x in [0, 1],

lim r F(t, Yn(t)) dt = r F(t, y(t)) dt.n~ooJo Jo

(See Problem 6, Section 6-1.)

ANSWERS

1. y/ + 2y = 0, y(l) = 12. y" + xy/ + 2y = 2, yeO)

3. (a) vex) = 2x - 2 + 2e-x

1(b) Vex) = ix + 3x2

n

4. Yn(X) = L: xk /k!

k=O

1, y/(O) = 0

n-l

5. Yn(X) = L: xk/k! (n ~ 1)

k=O

) + 1 n+l ( ) ( 26. Yn(X = 1 x - (n + I)! x 7. y x = 1 - x)

6-3 Proof of existence and uniqueness. Theorems 1, 2, and 3 of thissection verify the conditions I, II, III of the preceding section and thusconstitute a proof of the existence of a solution of

y/(x) = F(x, y(x)), yea) = b, (1)

- or equivalently

y(x) = b + LX F(t, Yet)) dt. (2)

The uniqueness of the solution is proved in Theorem 4, and the results ofthe chapter are summarized in Theorem 5 (Picard's theorem).

148

b+h

b

b-h

PICARD'S EXISTENCE THEOREM

I- 10 '1a-h a-ho a a+ho a+h

I· I ·1FIGURE 6-2

[CHAP. 6

The notation introduced next and illustrated in Fig. 6-2 is used throughout this section.

LetS = {(x,y): Ix - al ::s; hand Iy - bl ::s; hI·

Let [ = {x: Ix - al ::s; hI·

We will always assume that the function F in (1) and (2) is continuouson S, and it follows that F is bounded on S. Assume henceforth thatIF(x, y) I < M for (x, y) in S.

Draw the lines through (a, b) with slopes ±M, and let T be the shadeddouble-triangle region between these lines and within S.

Let [0 be the interval around a which is the projection of T on the x-axis.Let ho(ho ::s; h) be the half-length of [0, so that

[0 = {x: Ix - al ::s; hoIand

T = {(x, y) : x in [0 and Iy - bl ::s; M Ix - all. (3)

We recall the inductive definition of the approximating sequence {Ynl.

Yo(x) = b,

Y"n+l(X) = b + fax F(t, Yn(t») dt.(4)


THEOREM 1 (Condition I). If F is continuous on S, and IF(x,Y)1 < Mfor (x, y) in S, and 10, T are as indicated above, and IYn I is the sequencedefined by (4), then each function Yn is defined on 10 , and has its graph in T.

Proof. The proof is inductive. The function Yo obviously has its graphover 10 in T, so Yl is well defined on 10 by (4). Also,

IF(t, yo(t»1 = IF(t, b)j < M

for all tin 10 . For x in 10 ,

IYl(X) - bl = I.f F(l, Yo(t» dtl

~ IfMdtl

= Mix - aj. (5)

(6)

The inequality (5) is just the condition (3) that (x, Yl (x» be in T for xin 10 . Now suppose that above 10 the graph of Yn(n ;::: 1) is in T. Inparticular, F is continuous on the graph of Yn and so Yn+l is defined for xin 10 . Also IF(t, Yn(t» I < M for t in 10 , and hence

IYn+l(X) - bl = I.f F(t, Yn(t» dtl

~ Mix - al·

Thus Yn+ 1 also has its graph in T by (3). This completes the proof byinduction.

COROLLARY. Under the hypotheses of Theorem 1,

IYm(x) - Yn(x)/ ~ 2Mlx - al

for all m and n and all x in 10 .

Proof. Problem 1.

(7)

We have shown that the scheme (4) does define a sequence of functionson some fixed interval 10 around a. We will need an additional assumptionon F to show that the sequence converges.

DEFINITION. The function F satisfies a Lipschitz condition on S ifthere is a number A such that

(8)

for all points (x, Yl) and (x, Y2) in S.


The Lipschitz condition can be interpreted geometrically by saying thatalong any given vertical line (any fixed x), the function F(x, y), consideredas a function of the single variable y, nowhere increases more rapidly thanthe linear function Ay. This is true, for example, of the function F(x, y) =x 2 + xy on any region on which x is bounded. On the other hand, thefunction F(x, y) = 2"yxy (cf. Section 1-3) does not satisfy a Lipschitzcondition in any region around (1,0). Here Formula (8) would require, inparticular (x = 1), that

IF(1, y) - F(I, 0)1 = 12~1 ::::; AIY - 01 = AIYIfor all y in some interval around zero. This would say that for some constant A and all small values of y,

1 2~1 = 2-2/3 < Ay y -,

which is impossible.

THEOREM 2 (Condition II). IfF is continuous on S, and IF(x, Y)I <Mfor (x, y) ~'n S and F satisfies the Lipschitz condition (8), then the sequence{Yn I defined by (4) converges uniformly on 10 ,

Proof. We will consider instead of the sequence {Yn I, the series whosenth partial sum, for each n, is Yn. The uniform convergence of the sequence{Yn I is by definition equivalent to the uniform convergence of this series.We show that the terms of the series are smaller than the correspondingterms of a uniformly convergent power series, and thus complete the proof.

Recall the identity (cf. Problem 7, Section 6-1)

Yn(x) = Yo(x) + [Yl(X) - Yo(x)] +... + [Yn(x) - Yn-l(X)]. (9)

The sequence {Yn I is the sequence of partial sums of the series

00

Yo(x) + L: [Yn+l(X) - Yn(x)].n=O

We will show inductively that

(10)

(11)

for all x in 10 , The terms on the right of (11) are the terms of the series

(12)


The series (12) converges uniformly on every finite interval-in particularon Io-and hence the inequality (11) shows that (10) converges uniformlyon 10 ,

To prove (11), first consider n = 2:

!Y2(X) - Yl(X)/ = I.f; [F(t, Yl(t») - F(t, Yo(t»)] dtl

~ I..CAIYl(t) - Yo(t)1 dtl

~ Ii'" A2Mlt - al dtl

2MAlx - al 2= 2! .

In (13) we used the Lipschitz condition (8) to show that

jF(t, Yl(t») - F(t, Yo(t»)1 ~ AIYl(t) - Yo(t)1

and (7) to see that

!Yl(t) - Yo(t) I ~ 2Mlt - al·

Now suppose that (11) holds for some n ~ 2. Then

IYn+l(X) - Yn(X) I = Ii'" [F(t;Yn(t») - F(t,Yn_l(t»)]dtl

~ Ii'" AIYn(t) - Yn-l(t)1 dtl

11'" 2MAn-llt - aln I~ A I dt

a n.

2MAnix - aln +l

(n + I)!

(13)

(14)

Thus if (11) holds for n, it holds for n + 1 and, by induction, the inequalityholds for all n ~ 2. Since the series (12) converges uniformly on 10 , thesmaller series (10) converges uniformly on 10 , which is the same as sayingthe approximating sequence (YnJ converges uniformly on 10 ,

COROLLARY. If Y is the limit on 10 of (Yn J then Y is continuous on 10

and has its graph in T.

Proof. Problem 2.


THEOREM 3 (Condition III). If F is continuous on S and IF(x, y)1 < Mfor (x, y) in S, and F satisfies the Lipschitz condition (8) on S, and Y is thelimit of {Yn} on 10 , then

l~ f F(t, Yn(t» dt = f F(t, y(t» dt

for all x in 10 ,

Proof. The equality above can be written

By (8) we have

If IF(t, Yn(t» - F(t, y(t»1 dtl

~ If AIYn(t) - y(t)1 dtl

~ Alx - al max IYn(t) - y(t)1. tin 10

~ Aho max IYn(t) - y(t)l·tin 10

(15)

(16)

(17)

Since Yn converges uniformly on 10 to Y, for any E > 0 we can find N suchthat

Emax IYn(t) - y(t)1 < Ah

otin 10

for all n 2: N. Hence for n 2: N,

Ii'" [F(t,Yn(t» - F(t,y(t»ldtl < E

which verifies (16) and completes the proof.Theorems 1, 2, and 3, together with our observations of Section 6-2,

complete the proof of the existence of a solution. We proceed to the proofthat the solution is unique.

The following lemma shows that any solution of (1) or (2) has its graphin T. We will need this in Theorem 4 (proof of uniqueness) so we canapply the Lipschitz condition to any solution g, whether it arises as thelimit of a Picard sequence or not.

LEMMA 1. If F is continuous on S, and IF(x, y) I < M for (x, y) in S,and g is a solution of (1) or (2) on 10 , then the graph of g is in T.


s

Contradiction:g'(xo) =F(xo, y(xo») =M

butF(xo, y(xo») < M

1----10 -----..·1

FIG. 6-3. The graph of a solution lies in T.

Proof. If g is a solution on 10 , then g/ (a) = F(a, b) and hence Ig' (a) I < M.It follows (Problem 4) that the graph of g starts into T; that is, that

Ig(x) - bl < Mix - al

for all x in some interval (a - 0, a + 0). If the graph of g leaves T, sayto the right of a, then by the continuity of g (Problem 5) there will be asmallest number c > a where the graph crosses one of the diagonal lines(Fig. 6-3). That is, c will be the first number to the right of a such that

Ig(c) - bj = Mlc - al·

By the Mean Value Theorem, there is a number Xo in (a, c) such that

Ig(c) - g(a)1 = Ig/(xo)llc - al·

By the definition of c, (xo, g(xo) is in T, and hence

From (19), we conclude that

Ig(e) - bl < Mlc - aj

which contradicts (18). Therefore the graph of g cannot leave T.

(18)

(19)


THEOREM 4 (Uniqueness of the solution). If F' is continuous on S andIF(x, Y)I < M for (x, y) in S, and F satisfies the Lipschitz condition (8)on S and y, g are any two solutions of (2) over the interval 10 , then y(x) =g(x) for all x in 10 ,

Proof. The hypothesis is that y(a) = g(a) = b, and

y/(x) = F(x, y(x),

g/(x) = F(x, g(x),

for all x in 10 , We show that if y and g have the same value at any pointXo if 10 , then y(x) = g(x) on the interval [xo - 1/(2A), Xo + 1/(2A)]. Bystarting at Xo = a, we get y(x) = g(x) on [a - 1/(2A), a + 1/(2A)].Then taking Xo = a + 1/(2A), we get y(x) = g(x) on [a, a + 2/(2A)].With a finite number of repetitions of this argument, we show that y(x) =g(x) on all of 10 ,

Assume, therefore, that y(xo) = g(xo) for some Xo in 10 , and let h(x) =y(x) - g(x). We have h(xo) = 0, and we must show that h(x) = 0 on[xo - 1/(2A), Xo + 1/(2A)]. By Lemma 1, and (8), we have

Ih'(x)1 = Iy/(x) - g/(x)1

= IF(x, y(x) - F(x, g(x)1

~ Aly(x) - g(x)1

= Alh(x)l· (20)

Since h is continuous on [xo - 1/(2A), Xo + 1/(2A)], there is a point ofthis interval where Ih(x) I assumes its maximum value-say Ih(Xl) I is thismaximum value. By the Law of the Mean and the fact that h(xo) = 0,we have

Ih(Xl) I = Ih(Xl) - h(xo) I= Ih'Wllxl - xol,

for some number ~ between Xl and xo. From (20) and (21) we get

/h(Xl)1 = Ih'Wllxl - xol~ AlhWllxl - xol

1~ Alh(Xl)1 2A

= tlh(Xl)['

(21)

The inequality Ih(Xl)1 S tlh(Xl)1 implies that the maximum value/h(Xl)! = 0, and hence that h(x) == 0 on [xo - 1/(2A), Xo + 1/(2A)].


We recapitulate the results of the preceding theorems in Theorem 5.

THEOREM 5 (Picard's Theorem). If F is cont£nuous on a square S withcenter (a, b), and F satisfies (8) on S, then there is an interval 10 around asuch that there £s, on 10 , exactly one solution of (1).

COROLLARY. If F and F 71 are cont£nuous on a square S with center (a, b),then on some interval 10 around a there is one and only one solution of (1).

Proof. The conclusion of the corollary is merely a paraphrase of theconclusion of the Theorem. To prove the corollary is clearly sufficient toshow that the continuity of F71 implies that the Lipschitz condition (8)holds. The student is asked to carry out the details in Problem 7.

PROBLEMS

1. Prove the corollary to Theorem 1.2. Prove the corollary to Theorem 2. [Hint: Give a reason why each Yn is

continuous and cite the appropriate theorem from Section 6-1.]3. Prove that the integral on the right side of (15) exists.4. Assume the hypotheses of Lemma 1. Let f = M - IF(a, b)1 > O. There

is (why?) a positive number 0 such that

Ig(X) - b - F(a, b)! < M - IF(a, b)1x-a

if 0 < Ix - al < o. Show that this implies, as stated in the proof of Lemma 1,that Ig(x) ~ bl ~ Mix - al if Ix - al < o. [Hint: IA - BI < C is equivalentto B - C < A < B + C, -M ~ F(a, b) - f, and -D ~ A ~ D is equivalent to IAI ~ D.]

5. In the proof of Lemma 1, it is stated that g is continuous. Why is g continuous on 10?

6. Show that y(x) = x2 is not a solution on [0, 1] of any equation y' = F(x, y)with IF(x, Y)I < 1 for 0 ~ x ~ 1, 0 ~ Y ~ 1.

7. Prove the corollary of Theorem 5. [Hint: By its continuity, F lI must bebounded on S; say iFlI(x, Y)I < A for (x, y) in S. Use the Mean Value Theoremfor the function f(y) = F(x, y), x any fixed number in 1 (Fig. 6-2), to verifythat (8) holds on S.)

8. Show by the methods of this section that the equation y' = 1/(1 + y2),y(O) = 1, has a solution on every interval 10 = [-ho, hoL and hence a solutionon the whole line.

9. Show that [-!, !) is the biggest interval around zero on which the methodsof this section will guarantee a solution to y' = 1 + y2, y(O) = O. [Hint: Showthat the optimum square S to start with has sides of length 2 (that is, h = 1).)


6-4 Approximations to solutions. In many of the applications ofdifferential equations, the primary interest is in finding numerical valuesof a given solution. Most numerical statements are statements of approximation (e.g., 7r = 3.1416, V2 = 1.414, sin 7r/4 = .707), and all of themeasured quantities which the scientist and engineer are concerned withare approximations. For most numerical applications, therefore, there isno useful distinction between an approximate solution and an exact solution, provided the approximation is sufficiently good. It should be notedalso that even when a differential equation can be solved exactly, thenumerical values of the solution may not be readily accessible. For example, the equation

y3 + y = x (1)

characterizes the solution of

(2)y(O) = 0,, 1

y = 3y 2 + l'

but the values of this solution y cannot be obtained from (1) withoutfurther computation. Similarly, the solution of the linear equation

y' = 1 - 2xy, y(O) = 0 (3)

can be written2 r' 2y = e-X J

oet dt, (4)

but the Formula (4) by itself does not immediately yield numerical valuesof y.

The Picard proof not only shows that there is a solution of the equation

y' = F(x, y), y(a) = b, (5)

but gives a theoretical method of calculating approximations to the solution. Usually, however, the successive integrals which define the approximating functions become unmanageable, and the Picard method is not aneffective way of obtaining approximations. The first proof of an existencetheorem for (5), due primarily to Cauchy, is not so elementary as Picard'sproof, but it does provide an effective way of finding approximations to thesolution. Cauchy's method of proof, like Picard's, consists in defining asequence of approximate solutions, showing that this sequence convergeson some interval, and proving that the limit function must be a solution.The approximating functions of the Cauchy method are polygonal curvesobtained by following the direction field along short segments. The calculations involved in finding these approximations are purely arithmetic,and therefore well adapted to machine computation.

6-4] APPROXIMATIONS TO SOLUTIONS 157

To construct a polygonal approximation to the solution of (5) on aninterval [a, c], we proceed as follows. Let {xo, Xl, ... , xn} be a partitionof [a, c]; that is, let a = Xo < Xl < X2 < ... < Xn = c. ComputeF(a, b), and follow the line through (a, b) with slope F(a, b) until it intersects the line X = XI, say at the point (Xl, YI). Then follow the line through(Xl, YI) with slope F(xI, YI) until it intersects the line X = X2' Continuingin this way, we obtain a sequence of points (a, b), (Xl, YI), ... , (xn, Yn);the broken line path joining these points is the polygonal approximation forthe partition {xo, XI, . • . , xn}. If we let Yo = b, then we can write

k = 0, 1, ... , n - 1.

Since Yn = Yo + [YI - Yo] + ... + [Yn - Yn-tl, we have also

n-lYn = Yo + L F(Xk, Yk)(Xk+1 - Xk).

k=O(6)

A polygonal approximation to the solution on an interval [d, a] tothe left of the initial point can be constructed in a similar way.If d = Xn < Xn-l < ... < Xl < Xo = a, then the vertices (Xk, Yk) ofthe approximation are determined by

k = 0, 1, ... , n - 1,

and again we have

Yk+l = Yk + F(Xk' Yk)(Xk+1 - Xk).

In general, one gets better approximations by taking finer partitions, andfor any partition the approximation is likely to be best close to a.

For the sake of completeness we will state without proof* a theoremwhich indicates how polygonal approximations can be used to show theexistence of a solution of (5).

THEOREM 1. Assume F satisfies the hypotheses of Section 6-3. Let Pn bea sequence of partitions of 10 = [a - ho, a + ho] such that if IIPnIl1's themaximum distance between pQ1'nts of Pn, then limn-->oo IIPnl1 = 0. If YP isthe polygonal approximation for Pn, then the sequence {yP } converges ;;niformlyon 10 , and the limit function is a solution of (5). n

The following example shows a tabulation of the computations involvedin finding a polygonal approximation. We will find two approximations

* For a proof and further discussion see, e.g., Hurewicz, Lectures on OrdinaryDifferential Equations (New York: John Wiley and Sons, Inc., 1958), pp. 1-12.


to the solution of (3) on the interval [0, 2], using two different partitionsof the interval.

EXAMPLE 1. y' = 1 -2xy, yeO) = O.

TABLE 1

Partition 10, 0.5, 1.0, 1.5, 2.0 I

Xk 0 0.5 1.0 1.5 2.0Yk 0 0.5 0.75 0.5 0.25

F(Xk, Yk) = 1 - 2XkYk 1 0.5 -0.5 -0.5l:1Yk = !(1 - 2XkYk) 0.5 0.25 -0.25 -0.25

Yk + l:1Yk = Yk+l 0.5 0.75 0.5 0.25

TABLE 2

Partition 10,0.25,0.5,0.75,1.0,1.25,1.5,1.75, 2.0}

Xk 0 0.25 0.50 0.75Yk 0 0.25 0.4688 0.6016

F(Xk, Yk) = 1 - 2XA;yk 1 0.875 0.5312 0.0976l:1Yk = t(l - 2XkYk) 0.25 0.2188 0.1328 0.0244

Yk + l:1Yk = Yk+l 0.25 0.4688 0.6016 0.6260

Xk 1.0 1.25 1.5 1.75 2.0Yk 0.6260 0.5630 0.4611 0.3653 0.2956

F(Xk, Yk) = 1 - 2XkYk -0.2520 -0.4075 -0.3833 -0.2789l:1Yk = HI - 2XA;yk) -0.0630 -0.1019 -0.0958 -0.0697

Yk + l:1Yk = Yk+l 0.5630 0.4611 0.3653 0.2956

The graphs of these two approximations are shown in Fig. 6-4. Althoughone cannot conclude directly from Theorem 1 that the polygonal approximations will converge on an interval as large as [0, 2], this can be shownby a more careful examination of the particular function F in this example.

It is not easy to make a precise estimate of the accuracy of an approximation. Suppose one requIres, for example, accuracy to two decimalplaces. The usual procedure is to compute approximations for finer andfiner partitions until two successive approximations agree to two decimalplaces. Since the error from rounding off decimals may well accumulatefrom step to step in computing an approximation, it is usually necessaryto carry several more decimal places in the calculations than the accuracydesired.

6-4] APPROXIMATIONS TO SOLUTIONS 159

1.0

0.5

./I".....

./V .........

~

./? --re-.~ '-.-OV -.......,

~ r-.....

/ r-. .......,

V .........;

/1/

0.5 1.0 1.5 2.0

FIG. 6-4. Polygonal approximations for y'

PROBLEMS

1 - 2xy, yeO) = o.

1. Find a polygonal approximation on [0, ~] of the solution to y' = 1 - 2xy,yeO) = O. Use first the partition {O, 0.1, 0.2, 0.3, 0.4, 0.5} and then the partition{O, 0.05, 0.1, 0.15, ... ,0.45,0.51.

2. Find polygonal approximations to the solution of y' = x + y2, yeO) = 0,on the interval [0, 1]. Partition the interval into subintervals of length 0.2 andthen into subintervals of length 0.1. Plot the two approximations on the samegraph.

3. Find and graph the polygonal approximation to the solution ofy' = Ij(x + y), y(I) = 1.

(a) Use the partition {I,Ll, 1.2, 1.3, 1.4, 1.5} of [1,1.5].(b) Use the partition {0.5, 0.6, 0.7, 0.8, 0.9, 1.0} of [0.5,1].4. Let YP be the polygonal approximation to the solution of y' = f(x), yea) = 0,

for the partition p = {a = xo, xl, ... ,Xn = c} of [a, c]. Show that yp(c)( = Yn)is a Riemann sum approximation to J~f(x) dx [cf. Formula (6)].

5. Compute an approximate value for eO. l by finding the polygonalapproximation to the solution of y' = y, yeO) = 1 for the partition{O, 0.01, 0.02, ... , 0.09, O.I} of [0,0.1]. Carry five decimal places. (To fourdecimal places, eO. l = 1.1052.)

6. Find the polygonal approximation on [1, 2] of the solution of y' 2 - yjx,y(l) = 2. Use

(a) the partition {I, 1.2, 1.4, ,1.8, 2.0}(b) the partition {I,Ll, 1.2, , 1.9, 2.0}.

Solve the equation and plot the solution and the two approximations on the samegraph.

7. The solution of the problem of Example 1 is analytic [cf. Formula (4)],and the Picard approximations are the partial sums of the power series for thesolution. Find the Picard approximations Yo, Yl, Y2, Y3, Y4, and evaluate Y4(0.5).Show that Y4(0.5) is accurate to within 0.000033. [Hint: The series is alternating,so the error is less than the first term omitted: 16(~)9j3·5·7· 9).) Also find


Y4(1), Y4(2), estimate the error and compare these values with the polygonalapproximations of Table 2.

8. Use the differential equation (2) and the partition 10, 0.1, 0.2, ... , 0.9, 1.0Ito obtain an approximate graph over [0,1] for the function y defined by (1). Letx = 1 and solve (1) for y to three decimal places by trial and error.

ANSWERS

1. 1st partition Yk: 0, 0.1, 0.198, 0.2901, 0.3727, 0.44292nd partition Yk: 0, 0.05, 0.0998, 0.1488, 0.1966, 0.2427, 0.2866, 0.3280,

0.3665, 0.4018, 0.43372. 1st partition Yk: 0, 0, 0.04, 0.1203, 0.2432, 0.4150

. 2nd partition Yk: 0, 0, 0.01, 0.0300, 0.0601, 0.1005, 0.1515, 0.2138,0.2884, 0.3767, 0.4809

3. (a) 1, 1.05, 1.097, 1.141, 1.182, 1.221(b) 1, 0.950, 0.896, 0.837, 0.772, 0.699

5. 1.10466. (a) 2, 2, 2.07, 2.17, 2.30, 2.44

(b) 2, 2, 2.02, 2.05, 2.09, 2.14, 2.20, 2.26, 2.33, 2.40, 2.47Solution: y = x + l/x

7. Y4(0.5) = 0.42441, max error 0.000033Y4(1) = 0.525, max error 0.017Y4(2) = -4.6, max error 8.7

8. 0, 0.1, 0.197, 0.287, 0.367, 0.438, 0.501, 0.558, 0.610, 0.657, 0.701;y(l) = 0.682 by trial

CHAPTER 7

SYSTEMS OF EQUATIONS

7-1 Geometric interpretation of a system. In this chapter we willtreat systems of several simultaneous differential equations in severalunknown functions. We first consider the case in which there are twounknown functions, y and z, and try to interpret the situation geometrically. The systems we want to consider are those of the form

y' = F(x, y, z)

z' = G(x, y, z)

(Y' = ~~),

(z' = :;).

(1)

A solution of (1) is a pair of functions, (f, g), which satisfy (1) identicallyon some interval. That is, (f, g) is a solution of (1) on the interval I iffor all x in I,

j'(x) = F(x,f(x), g(x)),

g'(x) = G(x,f(x), g(x)).(2)

It is convenient to indicate solutions of a system with a pair of equations;thus we write

y = f(x),

z = g(x)(3)

to indicate that (f, g) is a solution of (1).Now let us ask whether we can reasonably expect the system (1) to

have solutions. First consider the geometric interpretation of the equations(3). The graph of·y = f(x) in three dimensions is a cylindrical surfaceconsisting of the points covered by a line parallel to the z-axis moving alongthe curve y = f(x) in the xy-plane (see Fig. 7-1). Similarly, the graph ofthe equation z = g(x) is a cylinder consisting of lines parallel to they-axis. The graph of the pair of equations (3) is the curve which is theintersection of these two surfaces. Hence we can picture a solution ofthe system (1) as being a curve in space.

Now let us ask what the system (1) demands of a curve in order that itbe a solution. Recall that the line tangent to the curve (3) at a point

161

162 SYSTEMS OF EQUATIONS

z

FIGURE 7-1

y

[CHAP. 7

(x,!(x), g(x)) has direction numbers (1, f'(x) , g'(x)). Comparison with(2) and (1) shows that in effect the system (1) prescribes at each point(x, y, z) a tangent line for a solution curve; namely, the line through(x, y, z) with direction numbers (1, F(x, y, z), G(x, y, z)). Thus the system(1) describes a tangent field in space, just as the single equation y' =

F(x, y) describes a tangent field in the plane. It is true, as one wouldexpect, that if F and G are continuous, then there is a solution curvefor (1) through each point in space. In other words, starting at any pointthe tangent field described by (1) "directs" through space a curve whichrepresents a solution. We will prove an existence theorem to this effectin Section 7-4 under an additional assumption (a Lipschitz condition)which allows us also to prove that the solution curve through any pointis unique.

EXAMPLE 1. y' = y/ x, z' = x - yz.

The first equation does not involve z and can readily be solved by separating variables. Accordingly, we know that for y and z to satisfy thesystem, we must in particular have y = C1X. Therefore z must satisfy

z' = x - C1XZ,

Z' + C1XZ = x.

7-1] GEOMETRIC INTERPRETATION OF A SYSTEM 163

The function e1/ Z(c 1x

2) is an integrating factor for this linear equation,

so z is given by

(if CI ~ 0).

If CI = 0, so that y = 0, then z' = x, and z = !XZ+ Cz.

EXAMPLE 2. y' = y - z, z' = y - z.

Here both unknowns occur in both equations, so we cannot solve eitherequation by itself. However, y' = z' if (y, z) is a solution, so we must.have y - z = CI for any solution. Therefore y and z must satisfy

y' = CI,

Y = CIX + Cz,

z' = CI,

The argument above shows that the last equations are necessary for asolution, but not that they are sufficient. If we substitute the functionsgiven above in either equation of the system, we get

Therefore Cl, Cz, and Ca are not independent, and the solutions of thesystem are given, for example, by

EXAMPLE 3. y' = z + x, z' = 2z - y + xz.

Here we can use the technique of solving one equation for one unknownand substituting in the other. From the first equation, we have

and hence

z = y' - x,

z' = y" - 1.

(4)

Substitution of these values for z and z' in the second equation of thesystem gives

y" - 1 = 2(y' - x) - y + x 2,

(5)y" - 2y' + y = X

Z- 2x + 1.

164 SYSTEMS OF EQUATIONS [CHAP. 7


y = Cle'" + C2xe'" + x2 + 2x + 3. (6)

From (6) and (4) it follows that if (y, z) is a solution of the system, thenwe must have

y = cle'" + C2xe'" + x2 + 2x + 3,

z = cle'" + C2(X + l)e'" + x + 2.(7)

11. y' = z,z'=2e"'-y

13. y' = x2 + !z,z' = 3x2 + 6y + z

Conversely, it is easy to show (Problem 8) that if y is a solution of (5)and z satisfies (4), then (y, z) is a solution of the system. That is, all pairs(y, z) given by (7) are solutions of the system.

Let us make use· of the example above to note the connection betweenthe existence theorem for second order equations and that for first ordersystems. Since z determines y' and vice versa by (4), there is a uniquesolution of (5) for any initial conditions yea) = bo, y'(a) = bl if and onlyif there is a unique solution of the system for any initial conditionsyea) = bo, z(a) = bl . We will use this sort of argument in Section 7-5to prove the existence theorem for a single nth order equation, after havingproved an existence theorem for first order systems.

PROBLEMS

1. Show that each line in the tangent field of the system y' = y/x, z' = z/xpoints toward the origin. Describe the solution curves.

2. Describe the tangent field and solution curves of the system y' = -x/y,z' = O.

3. Find the lines through the origin (y = ax, z = bx) which are solutioncurves of the system y' = xy/z2, Z' = yz/2x2.

4. Find the curves of the form y = ax2, z = bx which are solutions of thesystem y' = (y + z2)/x, Z' = (3z2 - 2xz)/y.

5. Solve the system, and check your answer. y' = z + y/x, z' = (x + z)/x.6. Solve the system, and check your answer. y' = y + z, z' = y + z.

Give an example to show that the following is false: (y, z) is a solution if and onlyif y - z = c.

7. Solve the sys~em and check your answer. y' = 3x2 + y - z, z' = y - z.[Hint: Subtract the equations.]

8. Show that if y is a solution of (5), and z is given by (4), then (y, z) is asolution of the system of Example 3.

Solve the following systems.

9. y' = z, 10. y' = y - z,z' = y z' = x - 2y

12. y' = y + z + x,z' = 4y + z + x + 4e'"

7-2] OTHER INTERPRETATIONS OF A SYSTEM

ANSWERS

165

1. All lines through the origin except those in the yz-plane2. The solutions are the circles parallel to the xy-plane with centers on the

z-axis.3. y = 2x, z = x; y = 2x, z = -x4. y = xz; z = x; y = 4xz, z = 2x5. y = xZ (In Iczxl - 1) + ClX, z = x In Iczxj6. y = czezx + !Cl, z = czezx - !Cl7. y = x3 + :1x4 + CIX + cz, z = :1x4 + CIX + Cz - Cl9. y = clex+ cze-x, z = Clex - cze-x

10. y = cle-x+ czezx + !x - :1, z = 2CICx- czezx + !x - i

11. Y Cl cos X+ Cz sin x + eX, z = -Cl sin x + Cz cos x + eX12. y Cle-x+ cze3x - !- - ex, z = -2Cle-x+ 2cze3x + !- - x13. Y Clezx + cze-x - x + !, z = 6clezx - 3cze-x - 3 - 3xz

7-2 Other interpretations of a system. A system of two first orderequations has interpretations other than the geometric one given in thepreceding section. To see how such a system might arise in mechanics,consider a particle moving in the xy-plane and write x = x(t), y = yet)for the coordinates of the particle at time t. Suppose that we know thevelocity (speed and direction) that the particle must have if it passesthrough the point (x, y) at the time t. That is, we assume that we knowthe x-component dx/dt and the y-component dy/dt of the velocity in termsof x, y, and t. This gives· us equations of the form

dxdt = F(t, x, y),

dydt = G(t, x, y).

(1)

This system is identical to system (1) of the preceding section, except wehave relabeled the independent variable t, and the unknown functionsx and y.

Although the situation outlined above does occur, it is not typical inproblems of motion. One is more likely to know the forces on a particle,in terms of position, time, i'tnd velocity, than to know the velocity interms of position and time. In this case, the motion is described by thesystem of two second order equations obtained by equating m(dzx/dtZ)and m(dZy/dtZ) to the horizontal and vertical forces.

Another interpretation of (1) results if we regard t simply as a parameter,and a solution

x = f(t), y = get)

of (1) as being the parametric equations of a plane curve. For example,


-a

H

A

(0, a)

a

y=acosh~a

2a

FIG.·7-2. The catenary.

in setting up the differential equation of a curve it may be easier to determine the components dx/dt and dy/dt of the tangent vector in terms ofa parameter t than it is to find dy/dx in terms of x and y. The followingexample illustrates this procedure of describing a plane curve by a pairof differential equations involving a parameter.

EXAMPLE 1. Find the equation of the curve (called a catenary) .formedby a flexible cable suspended from two points (Fig. 7-2).

Let the linear density of the cable be w, and let the tension in the cableat the low point (A) be H. Define a by the equation H = aw. It is convenient to introduce coordinates so that the coordinates of the point Aare (0, a). We will find the coordinates (x, y) of any point B on the curvein terms of the length s of the arc from A to B.

Let T be the tension in the cable at B. The arc from A to B is inequilibrium, so the horizontal component of T must equal the horizontalcomponent H of the tension at A. Also, the vertical component Vof Tmust equal the weight ws of the arc AB. Since dx/ds and dy/ds are thecomponents of a unit tangent vector at B, it follows that T (dx/ds) = H,and T (dy/ds) = V. We have the following formulas:

dx H dy Vds = T' ds = T' (V = W8, H = wa),

T = VH2 + V2 = wva2 + 82•

The system of equations which describe the curve is

dxd8

a dyd8

8

(2)

7-2] OTHER INTERPRETATIONS OF A SYSTEM 167

Integration of these equations gives

x = a sinh-1 ~,a

(3)

where both constants of integration are zero because of the way the coordinates were chosen. The parameter s can be eliminated (Problem 2) fromequations (3) to give

y = a cosh ~.a

PROBLEMS

(4)

1. Solve the system (2) and obtain (3). [Hint: Use the substitution 8 =a sinh () to integrate the first equation.]

2. Eliminate the parameter 8 from equations (3) to obtain (4). [Hint: Solveboth equations for 82 .]

3. In Example 1, the arc length 8 of the curve from A(O, a) to B(x, y) is givenby 8 = fo Y!1 + y'2 dx (y' = dy/dx). Show that y' = V /H = 8/a and thatthe curve can be described by ay' = fo Y!1 + y'2 dx. Differentiate this equationand solve the resulting second order equation.

4. A curve through (0, 1) is such that the slope of the tangent line at any point(x, y), x 2: 0, is equal to twice the length of the arc from (0, 1) to (x, y). Findthe equation of the curve. Find the coordinates of the point on the curve whosedistance from (0, 1) along the curve is one unit.

and dy/ ds = dy = 28 ]dx/d8 dx .

5. A marble is propelled up in a vacuum at an angle a from the horizontal,with initial velocity v. Find the position at time t and show the path is parabolic.

6. A bomb of mass m is dropped from an airplane flying horizontally at speed v.The horizontal and vertical components of the force of air resistance are respectively a(dx/dt)2 and a(dy/dt)2, where a is constant. Take the origin as the droppoint, with the positive y-axis pointing down, and show that the trajectory isdescribed by the system

d2x

= -: (~:Y (x 0,dx

when t = O}dt2

- = vdt

d2y a (dYY

(ydy

= ° when t = O}dt2 = g -:;:;; dt dt

Find x and y at time t.7. An object slightly heavier than water is released under water by a sub

marine moving horizontally at small velocity v (so there is nonturbulent flowthrough a somewhat viscous medium). The force of water resistance in this casehas components a(dx/dt) and a(dy/dt). If the object has mass m, and weightin water W, find the parametric equations of the path.


ANSWERS

[CHAP. 7

3. y = a cosh (~)

4. x = t sinh-1 2, y = ~ + v:: ;y = HI + cosh 2x)

5. x = (v cos a)t, Y = (v sin a)t - !gt2

6. x = ~ln[~vt+ 1]' Y = ~lncosh(~t)7. x = mv (1 _ e-<alm)!), y = Wm (e-<alm)! _ 1) + W t

a ~ a

7-3 A system equivalent to M(x, y) dx + N(x, y) dy = O. We haveseen that the solutions of the system

dxdt = F(t, x, y),

(1)dydt = G(t, x, y)

can be interpreted as a family of plane curves given in parametric form.Weare familiar with the fact that the solutions of

M(x, y) dx + N(x, y) dy = 0 (2)

form a family of curves. It is natural to ask whether one can always finda parametric system (1) which has the same solution curves as any givenequation -(2). If so, we would expect the system in some cases to beeasier to solve than (2), since many curves are more simply describedparametrically than by an equation in x and y. We can in fact show thatthe equation (2) is equivalent to the system

dxN(x, y)

which we agree is the same as

dy = dt-M(x, y)

dxdt = N(x, y),

(3)

dy M(-dt = - x, y).

Here "equivalent" means that every solution curve y = hex) of (2) hasa parametric representation x = J(t), Y = get) which constitutes a solu-

7-3] SYSTEM EQUIVALENT TO M(x. y) dx + N(x, y) dy = 0 169

tion of (3), and vice versa. Formally, this says that if x = f(t), Y = g(t) isa solution of (3), and h is a function such that g(t) == h(j(t»), theny = h(x) is a solution of (2). Conversely, if y = h(x) is a solution of (2),then there is a solution x = f(t), Y = g(t) of (3) such that g(t) == h(j(t»).These facts are proved in the following theorem.

THEOREM 1. If f and g satisfy

f'(t) == N(j(t), g(t»),

g'(t) == -M(j(t), g(t»),

and

g(t) == h(j(t»),

then h satisfies

M(x, h(x») + N(x, h(x)h'(x) == O.

(4)

(5)

Conversely, if h satisfies (5), then there are functions f and g satisfying (4)such that g(t) == h(j(t»).

Proof. The first part of the proof is left to the student (Problem 1).Let us show that every solution h of (2) has a parametric representationsatisfying (4). Assume therefore that h satisfies the identity (5). Let fbe a solution of the differential equation

f'(t) = N(j(t»), h(j(t»). (6)

Define g by the equation g(t) = h(j(t»). With this definition of g, (6) isthe first of equations (4), and we need only verify the second. From thechain rule we have g'(t) = h'(j(t»)f'(t). Now rewrite the identity (5) bysubstituting f(t) for x; by definition of g we can then write g(t) for h(x) =h(j(t»), and g'(t)/f'(t) for h'(x) = h'(j(t»). The result is

g' (t)M(j(t), g(t») + N(j(t), g(t») f'(t) == O.

By rearranging and comparing with (6), we get

f'(t) _ g'(t) == 1.

N(j(t), g(t») -M(j(t), g(t»)

Therefore g satisfies the second of equations (4). We have found a solution (f, g) of (3) such that x = f(t), Y = g(t) is a parametric representationof the curve y = h(x).


EXAMPLE 1. (2x - y) dx + VI - X 2 dy = O. The equivalent systemcan be written

or

dx dy = dt2x - y ,

dydt = Y - 2x. (7)

The variables separate in the first equation, and we can integrate to get

x = sin (t + Cl)'

Substitution in the second equation gives

dydt - Y = -2 sin (t + Cl)'

There is a particular solution of this linear equation of the form y =A cos (t + Cl) + B sin (t + Cl), and we determine by substitution thatA = B = 1. The solutions of the system (7) are therefore

y = C2et + cos (t + Cl) + sin (t + Cl),

x = sin (t + Cl)'

Using the relations

cos (t + Cl) = VI - sin2 (t + Cl) = VI - x 2,

we can eliminate t to obtain

y = cesin-lx + vI - x 2 + x,

PROBLEMS

(8)

1. Prove that if f and g satisfy (4) and get) == h (j(t) ), then h satisfies (5).2. The equation of Example 1 can be written y' = (y - 2x)/vl - x2 •

For what points (a, b) is there a solution y such that yea) = b? Check, by substitution, that the functions (8) are solutions of the equation. Use the uniquenesstheorem (Theorem 4 of Section 6-3) to show that the family (8) contains allsolutions.

3. List the various techniques of Chapter 1 for solving a first order equationand see whether any of them apply to the equation of Example 1.

7-4] EXISTENCE AND UNIQUENESS THEOREMS 171

Solve Problems 4 through 6 by the methods of Chapter 1, and also by writingas a system. Reconcile the answers by eliminating the parameter from the solution of the system.

4. xy dx + e-'" dy = O.5. (e'" + y) dx + e-'" dy = O. [Hint: To integrate fee"'e2'" dx, let u = e"',

dx = (lju) du.]6. [-(x4 + y2)jx2]dx + (yjx) dy = O. [Hint: To solve directly, try the

substitution y = ux.]

Write an equivalent system and solve.

7. (2Y +Ddx + (x + 2y2) dy = O.

8. (y + sin-Ix) dx + VI - x2 dy = O.9. (y - 2x) dx + (1 + y) dy = O.

10. (2y - 2y2) dx + (2X - ; + 1) dy = 01 du 1

[Hint: Let u = -, - = -"2 (dyjdt).]y dt y

ANSWERS

2. Any point (a, b) with lal < 14. x = In (t + Cl), In Iyl = (t + cr)[1 - In (t + Cl)] + C2;

In Iyl = (1 - x)e'" + C

5. x = In (t + cr),y = c2e-t - t - cr + 1;y = ce-e'" - e'" + 1 (c = C2e-C1)

6. x = cret + c2e-t, y = x(clet - cze-t);y2 = X2(X2 + c), (c = -4ClC2)

7. 2y2 + 1 = cre4t, x = -!cle-4t - 1 + c2e-t

8. x = sin (t + cr), y = C2e-t + t - 1 + cr9. x = clet + c2e-2t - !, y = cret - 2c2e-2t - 1

10. y = (cle2t + 1) -1, X = (C2 - crt)e2t

7-4 Existence and uniqueness theorems. In this section we prove anexistence and uniqueness theorem for a pair of simultaneous first orderequations in two functions. The theorem states that for appropriate Fand G there is one and only one pair of functions (y, z), defined on someinterval around a, which satisfy the equations

y'(x) = F(x, y(x), z(x),

z' (x) = G(x, y(x), z(x),

yea) = b,

z(a) = c.(1)

The method of proof is the Picard method of Chapter 6, with only themodifications required to treat two equations simultaneously.


We first rewrite the system (1) as an equivalent pair of integral equations:

Next we define inductively two sequences IYn} and IZn} by the equations

y(x) = b +f F(t, yet), z(t) dt,

z(x) = c + f; F(t, yet), z(t) dt.

Yo(x) = b,

zo(x) = c,

Yl(X) = b +f' F(t, b, c) dt,

Zl(X) = c + f' G(t, b, c) dt,

Yn+l(X) = b + LX F(t, Yn(t), Zn(t) dt,

Zn+l(X) = C + LX G(t, Yn(t), Zn(t) dt.

(2)

(3)

I

I

I

I

I

I

I

I

The proof consists in showing that these sequences IYn} and Izn} convergeand that the limit functions constitute a solution. If there are limits Yand Z of the respective sequences IYn} and Izn} such that

l~ LX F(t, Yn(t), Zn(t) dt = f F(t, Yet), z(t)) dt,

lim r G(t, Yn(t), zn(t) dt = (X G(t, yet), z(t)) dt,n~ooJa Ja

(4)

it is clear that these limits satisfy (2), because of the defining relation (3).As in the proof of Picard's theorem for a single equation, there are three

things which must be checked to make a proof along the lines indicated.We must show that

I. The scheme (3) defines all Yn and Zn on some common interval10 around a.

II. The sequences IYn} and Izn} converge on 10 to some limits Y and z.III. The integrals in (4) converge as indicated for all x in 10 •

We state next the hypotheses on F and G under which we can verifythe conditions I, II, and III; these assumptions will be used throughoutthis section.

7-4) EXISTENCE AND UNIQUENESS THEOREMS 173

The functions F and G are assumed to be continuous on some cube Swith center at (a, b, c):

S = {(x, y, z) : Ix - al ~ h, Iy - bl ~ h, Iz - cl ~ hI.

Since F and G are continuous on the closed bounded set S, they arebounded on S. Let M be a number such that 1 ~ M, and IF(x, Y, z)1 < M,IG(x, Y, z)1 < M for all (x, Y, z) in S. Let ho = hiM, so that ho ~ h,and Mho = h. The interval 10 of conditions I, II, and III will be[a - ho, a + hol. We assume that F and G are not only continuous, butalso satisfy the following Lipschitz condition on S: for some number Aand any two points (x, Y, z) and (x, Yb Zl) in S,

IF(x,y,z) - F(X'Yl,Zl)1 ~ A{ly - Yll + Iz - zllL

IG(x, Y, z) - G(x, Yb zl)1 ~ A{ly - Yll- Iz - zdl·

Now we proceed with the several parts of the proof.

(5)

IYn+l(X) - bl = ,lax F(t, Yn(t), zn(t)) dtl

~ Mix - aJ ~ Mho = h.

THEOREM 1 (Condition I). For each n, the functions Yn and Zn given by(3) are defined on 10 , and for every x tOn 10 , the point (x, Yn(x), zn(X))is in S.

Proof. (By induction.) The functions Yo and Zo are obviously definedon 10 , and the points (x, Yo(x), zo(x)) = (x, b, c) are in S if x is in 10 .

Suppose that Yn and Zn are defined on 10 for some n, and that(x, Yn(x), zn(x)) is in S for all x in 10 . Since F and G are continuous onthe points (t, Yn(t), Zn(t)), tin 10 , the integrals in (3) defining Yn+l(X) andZn+l(X) exist for all x in 10 ; that is, Yn+l and Zn+l are defined on 10 .

Using the fact that W(x, Y, z)1 < M for (x, Y, z) in S, we have, for everyx in 10 ,

Similarly,

for all x in 10 , and the point

is in S.

THEOREM 2 (Condition II). The sequences {Ynl and {znl converge uniformly on 10 .


Proof. We will actually demonstrate the uniform convergence of thetwo series whose respective nth partial sums are Yn and Zn. That is, wewrite

Yn(X) = Yo(x) + [YI(X) - Yo(x)] + ... + [Yn(x) - Yn-I(X)],(6)

Zn(X) = ZO(X) + [ZI(X) - ZO(X)] + ... + [Zn(X) - Zn_I(X)]

and prove that the following series converge uniformly on 10 ,

co

L: [Yn+I(X) - Yn(X)];n=l

(7)co

L: [Zn+I(X) - Zn(X)].n=l

The Lipschitz condition (5) is used to estimate the size of the terms of theseries (7) as follows:

IY2(X) - YI(X)! = If' [F(t, YI(t), Zl(t») - F(t, yo(t), zo(t»)] dtl

::; If A{IYI(t) - Yo(t)1 + IZI(t) - zo(t)l} dtl

::; IL~ A{h + h} dtl

= 2Ahlx - al.

The same argument on G shows that

Using the inequalities just proved, we estimate IY3(X) - Y2(x)1 andIZ3(X) - z2(~)I:

IY3(X) - Y2(X)!. = If' [F(t, Y2(t), Z2(t») - F(t, YI(t), Zl(t»)] dtl

::; Ii'" A{IY2(t) - YI(t)! + !Z2(t) - zl(t)l} dtl

::; Ii'" A{4Ahlt - al} dtl

= 4~;h Ix _ a1 2•

Again the same argument on G shows that

7-4] EXISTENCE AND UNIQUENESS THEOREMS 175

Continuing in this way, we see that for every n and every x in 10 , we have

(8)

Since the series :E:=l BnIn! converges for every number B, the estimates(8) show that both series (7) converge uniformly on 10 . This is the sameas saying that the sequences of partial sums IYn} and IZn} convergeuniformly on 10 ,

THEOREM 3 (Condition III). If Y and Z are the limits of the sequencesIYn} and Izn} defined by (3), then for all x in 10

l~ f; F(l, Yn(t), Zn(t» dt = f F(t, y(t), z(t» dt,

and

l~ {' G(t, Yn(t), zn(t» dt = f G(t, Yet), z(t» dt.

Proof. We treat the first equation above, writing it in the form

l~ faX [F(t, Yn(t), Zn(t» - F(t, Yet), z(t»] dt = O. (9)

It is sufficient to show that the integrand in (9) can be made uniformlysmall on 10 by taking n sufficiently large. From the Lipschitz condition(5), we get

IF(t, Yn(t), Zn(t» - F(t, yet), z(t»1

~ AIIYn(t) - y(t)1 + IZn(t) - z(t)I}·

Since the sequences IYn} and Izn} converge uniformly on 10 to Y and z,the right side above will be uniformly small on 10 for all sufficiently large n.The second equality of the theorem follows from the same argument applied to G.

This completes the proof that there is a solution (y, z) of (1) on theinterval 10 , Notice that we can specify a minimum length for 10 , interms of the bound M for F and G on S, and that we can characterizethe functions Y and Z as limits of constructable sequences. It remains tobe shown that the pair of functions found above is the only solution of (1)on 10 ,

In order to apply the Lipschitz condition in the uniqueness proof, weneed the following lemma, which states that any solution curve of (1)must stay within S, for x in 10 ,


LEMMA 1. If (YI, Zl) is any solution of (1) or (2) on 10 , then(x, YI(X), ZI(X) is in S for all x in 10 .

Proof. If (x, YI (x), Zl (x) is not in S for some x in Io-say some x > a-then by the continuity of YI and ZlJ there is a smallest number Xlgreater than a such that (Xl, YI (Xl), Zl (Xl) is on the boundary of S.That is, IYI (Xl) - bl = h or IZI (Xl) - cl = h. To be specific, supposethat IYI(XI) - bl = h and for all X between a and Xl, !YI(X) - bl < hand IZI(X) - cl < h. By the Mean Value Theorem, there is a number Xobetween a and Xl such that

Since (Yb zd is assumed to be a solution of the system (1), we have, inparticular,

yi(xo) = F(xo, YI(XO), ZI(XO),

and hence ly{(xo)1 < M. Therefore, from (10), we get

In other words, the contradiction h < h follows from the assumption that(x, YI(X), ZI(X) lies outside S for some X in 10 .

THEOREM 4 (Uniqueness of the ~olution). If (y, z) and (Yb Zl) are anysolutions on 10 of (1) or (2), then Y = YI and Z = ZIon 10 .

Proof. The proof is essentially the same as the proof of uniqueness fora single equation (cf. Theorem 4, Section 6-3). We show that if y(xo) =YI (xo) and z(xo) = Zl (xo) for some point Xo in 10 , then Y = YI andZ = Zl and the interval [xo - 1/(2A), Xo + 1/(2A)], where A is the constant of the Lipschitz condition. Since yea) = YI(a) = b, and z(a) =

Zl (a) = c, we can start the argument at Xo = a, and repeat it a finitenumber of times to show that Y = YI and Z = zIon all of 10 •

Assume that y(xo) = YI (xo), and z(xo) = Zl (xo) for some Xo in 10 •

Let hex) = vex) - YI(X), and k(x) = z(x) - ZI(X), so that h(xo) =k(xo) = o. We must show that hex) = 0 and k(x) = 0 for all X in[xo - 1/(2A), Xo + 1/(2A)]. By Lemma 1 and the Lipschitz conditionon F and G, we obtain

Ih'(x)! = Iy'(x) - yi(x) I= IF(x, Vex)~ - F(x, YI(x)1

~ Aly(x) - YI(X)!

= Alh(x)l, (11)

7-4]

and, similarly,

EXISTENCE AND UNIQUENESS THEOREMS

Ik'(x)1 ~ Alk(x)l·

177

(12)

Now let Ih(XI)1 and Ik(X2)1 be the maximum values that Ih(x)1 and Ik(x)1assume on the closed interval [xo - 1/(2A), Xo + 1/(2A)]. By the Lawof the Mean,

and

for some h between Xo and Xl, and some b between Xo and X2' Using(11) and (12) with the last equalities, and the fact that IXI - xol < 1/(2A)and IX2 - xol < 1/(2A), we get

Ih(XI) I ~ A Ih(XI) Ilxl - xol < !lh(XI) I,and

Hence Ih(XI)! = 0 and Ik(X2) I = 0, and since the maximum values ofIh(x)1 and Ik(x)1 are zero, hand k are identically zero on [xo - 1/(2A),Xo + 1/(2A)].

This completes the proof that the system (1), for suitable functions Fand G, has a unique solution on some interval around a. Our results arerestated, in slightly more convenient form, in the following theorem.

THEOREM 5 (Summary). If F, G, F II, Fz, Gil' Gz are continuous in somecube containing (a, b, c), t1J,en there is, on some interval around a, one andonly one solution (y, z) of the system

y' = F(x, y, z),

z' = G(x, y, z)

such, that y(a) = b and z(a) = c.

Proof. The continuity of the partial derivatives of Ii' and G impliesthat F and G satisfy the Lipschitz condition of the earlier theorems(Problem 5).

The system (1) is called a linear system if the functions F and G arelinear in y and z; i.e., a linear system is one of the form

y' = Pll(X)y + P12(X)Z + ql(X),(13)


Linear systems allow the same sort of systematic treatment as single linearequations and constitute the most important type of system. The existence theorem for single linear equations, which is the foundation for all ofChapters 3 and 4, will be proved on the basis of the existence theorem forlinear systems.

If Pll, Pl2, P21> P22, q1> and q2 are all continuous on some interval I,then the system (13) satisfies the hypotheses of Theorem 5 for any (a, b, c)with a in I. However, we can obtain a much better result for (13) than forthe general system (1). Theorem 5 says that (1) has a solution on someinterval around a; this interval may be small even if the functions F and Gand their partial derivatives are continuous everywhere (see Problem 7).For the linear system (13), we can show that there is a solution (y, z) defined on all of any interval on which the coefficient functions are continuous.

THEOREM 6. If Pll, Pl2, ql, P21> P22, and q2 are continuous on [a, 13),and a < a < 13, and b, c are any numbers, then there is one and only onesolution (y, z) of (13) on [a, I3l such that y(a) = b and z(a) = c.

Proof. The proof is as before, only now we can show that the sequencesIYn} and Izn} converge on the whole interval [a, 13). The integrals (3)defining Yn and Zn l:\.re clearly defined for all x in [a, 13). (Condition I).The rest of the proof of Theorem 5 was based only on the fact that thepoints (x, Yn(x), zn(x) were in S, where the Lipschitz condition on F andG was known to hold. For the linear case, the Lipschitz condition holdson any set of points (x, Y, z) such that x is in [a,l3) [Problem 8(b)). Theconvergence required in Conditions II and III now follows on all of [a, 13l,as in the proofs of Theorems 2 and 3. Th~ student is asked to supply thedetails of this argument in Problem 8.

PROBLEMS

1. Show that the system of integral equations (2) is equivalent to the system(1).

2. Solve the following system by finding Yn(x), zn(x) for all n, and verifyingconditions I, II, and III directly.

y' = z, z' = Y;

y(O) = z(O) = 1.

3. Find Y2 and Z2 for the system y' = z - 1, z' = Y - Xi y(l) = 0, z(l) = 2.4. Make a formal proof by induction of the inequalities (8).5. Prove Theorem 5 from the preceding theorems; i.e., show that the conti

nuity of FII, F., Gil' G. on S (and in particular the boundedness of these functions)implies that F and G satisfy (5).

I

I

I

I

I

I

7-5] EXISTENCE THEOREM FOR nTH ORDER EQUATIONS 179

6. (a) Show that the function G(x, y, z) = !z2/3 does not satisfy the Lipschitzcondition of (5) on any cube S containing (0,0,0). [Hint: If the condition holds on S, then !G(O, 0, z) - G(O, 0, O)I/Iz - 01 is boundedon S.] .

(b) Find two solutions of the system y' = yz, z' = !z2/3 such that yeO) =z(O) = O.

7. Show that no solution of the following system is defined on an intervallonger than 11'/10: y' = 100 +y2, Z' = y.

8. Assume that the functions Pll, Pl2, ql, P21, P22, q2 of the linear system(13) are continuous on [a, ~].

(a) Show that the sequences {Yn} and {Zn} given by (3) are defined on [a, {j].(b) Show that (5) is satisfied on {(x, y, z) : a ::; x ::; {j}.(c) Show that{Yn} and {Zn} converge uniformly on [a, {j].(d) Show that the limits y and z of the sequences {Yn} and {Zn} are solutions

of (13) on [a, {j].(e) Show that there is only one solution of (13) on [a, (j].9. Show that y is a solution of the second order equation y" = G(x, y, y') if

and only if (y, y') is a solution of the system y' = z, z' = G(x, y, z). State andprove an existence and uniqueness theorem for y" = G(x, y, y'), using Theorem 5.

10. State and prove an existence and uniqueness theorem for the linearequation y" + PlY' + Poy = q, using Theorem 6.

ANSWERS

2. Yn(X) = zn(x) = 1 + x + x2/2! + ... + xn/n! y(x) = z(x) = eX3. Y2(X) = 2(x - 1) - !(x3 - 1); Z2(X) = 2 - !(x2 - 1) - (x - 1)6. (b) y = 0, z = 0; y = 0, z = x3

7. All solutions must satisfy y = 10 tan (lOx + c),z = -In Icos (lOx + c)l, for some c.

7-5 Existence theorem for nth order equations.. In this section we willextend the results of Section 7-4 to first order systems in n functions, andexamine the connection between such systems and nth order equations inone function. The systems we consider are those of the form

Y) = Fl(x, Yll , Yn),

y~ = F 2 (x, Yll , Yn),(1)

A solution of (1) is an n-tuple of functions, (Yll ... , Yn), such that on someinterval the following identities hold:

k = 1, ... , n.

If the functions F I , ... , Fn are sufficiently well behaved near a point


(a, bl , ... ,bn ), then· there is, on some interval around a, exactly onesolution (Yl' ... , Yn) of (1) which satisfies the initial conditions

Yl (a) = bI, ... , Yn(a) = bn. (2)

Theorem 5 is a precise formulation of this statement for the case n = 2.The existence theorem was proved for the special case n = 2, rather thanfor the general system (1), so that the notation could be kept as simple aspossible, and the statements interpreted in familiar geometric terms.Except for the welter of subscripts required, the proof of the existenceand uniqueness theorem for (1) is not essentially different from that givenin Section 7-4. The approximating sequences {Ylml, ... , {Ynml are defined inductively as before by the formulas

YkO(X) = bk,(3)

Yk,m+l(X) = bk +f' Fk(t, Ylm(t), ... , Ynm(t») dt (k = 1,2, ... ,n).

By the same methods used in Section 7-4 these sequences can be shown toconverge on some interval around a to functions YI, ... , Yn which constitute a solution of (1) satisfying the initial conditions (2). We will notrepeat the details of the proof; the statement of the existence theoremfor (1) is as follows:

THEOREM 1. If FI, ... , Fn and the partial derivatives (iJjaYj)Fk(j, k = 1, ... , n) are continuous on the set of points (x, YI, ... , Yn) suchthat Ix - al ~ h, !Yl - bll ~ h, ... , IYn - bnl ~ h, then there is oneand only one solution (yI, ... , Yn) of (1) on some interval around a suchthat Yl(a) = bI, ... , Yn(a) = bn.

A linear first order system in n functions is a system of the form

y~ = Pll(X)Yl + + Pln(X)Yn + ql(X),

y~ = P2l(X)Yl + + P2n(X)Yn + q2(X),(4)

y~ = Pnl(X)Yl + ... + Pnn(x)Yn + qn(x),

The proof given in Section 7-4 for the existence of solutions to a linearsystem in two functions also extends without essential change to thegeneral case. We therefore have the following theorem for the linearsystem (4).

THEOREM 2. If the functions Pij, qi, for i, j = 1, ... , n, are continuouson [a, m, and a, bI, ... , bn are any numbers, with a < a < (3, then (4)


has a unique solution (Yl' ... , Yn) on [a,,8] such that Yl(a) = bl, ... ,Yn(a) = bn.

Using Theorem 1, we can now give a very simple proof of the existencetheorem for a single nth order equation in one function (cf. Problem 9,Section 7-4). To show that the equation

y(n) = F(x, Y, y', ... ,y(n-l») (5)

has solutions, we recast it as a first order system in the functions Y,y', ... ,y(n-ll. To conform with the notation of Theorem 1, we- definefunctions Y2, ... , Yn by

Y' = Y2, " (n-l)Y = Ya, ... , Y = Yn. (6)

With the agreement (6), Y is a solution of (5) if and only if (y, Y2, ... , Yn)is a solution of the system

Y' = Y2,

y~ = Ya,

y~ = F(x, Y, Y2, ... , Yn).

Initial conditions for (5)

(7)

y(a) = bo, y'(a) = bl, ... ,y(n-l)(a) = bn- 1 (8)

correspond to the following initial conditions for (7).

y(a) = bo, (9)

The first n - 1 functions on the right of (7)-the functions F 1, ... , Fn-lin the notation of Theorem I-satisfy the hypotheses of Theorem 1 forany point (a, bo, ... ,bn - 1). Therefore (7) has a solution satisfying theinitial conditions (9) provided F satisfies the condition of Theorem 1. Itfollows that (5) has a solution satisfying the initial conditions (8) if Fsatisfies this condition, and we have proved the following theorem.

THEOREM 3. If F and its partial derivatives F II, F II" ... ,FII(n-ll arecontinuous on the set of points (x, Y, y', ... , y(n-ll) such that Ix - al ~ h,Iy - bol ~ h, IY' - b1 ! ~ h, ... , Iy(n-ll - bn- 1 1 ~ h, then there is,on some interval around a, a unique solution Y of (5) satisfying the initialconditions (8).

Remark. There is a certain amount of notational confusion in thestatement of the above theorem. Although the symbols x, Y, y', y", ... ,

182 SYSTEMS OF EQUATIONS [CHAP, 7

y(n-ll are not customarily used as dummy coordinates for a point in(n + I)-dimensional space, they are used in this way in the statement ofthe theorem to conform with the appearance of F in Eq. (5). Thus F is afunction of n + 1 variables, and Fy', for example, means the partialderivative of F with respect to the third variable. The symbols y', y", etc.,in (8) refer as usual to the derivatives of the solution y.

The same sort of argument used to prove Theorem 3 from Theorem 1can be used to deduce Theorem 4 below from Theorem 2.

THEOREM 4. If Po, PI, ... , Pn-lI q are cont£n\uous on [a, {1], and a, bo,bI, ... , bn- l are any numbers with a < a < (1, then the linear equation

y(n) + Pn~I(X)y(n-l) + ... + PI(X)y' + Po(x)y = q(x)

has a unique solution y on [a, (1] such that yea) = bo, y'(a) = bll ... ,

y(n-ll(a) =, bn- l .

Proof. Problem 4.

Systems of higher order equations can also be treated as first ordersystems by introducing the derivatives as new functions. For example,the system

y' = F(x, y, z, z'),(10)

Z" = G(x, y, z, Z')

is equivalent to the first order system

Z' = w,

y' = F(x, y, z, w),

w' = G(x, y, z, w).

It follows that if F, G and their partial derivatives with respect to the lastthree variables are continuous. in a region containing the point (a, b, c, d),then there is a unique solution (y, z) of (10) on some interval around asuch that yea) = b, z(a) = c, and z'(a) = d.

PROBLEMS

1. Find the approximating functions YIO, Y20, Y30j ••• j Y13, Y23, Y33 as givenby (3) for the system

Y~ = Y2,Y~ = Y3,Y~ = -YI,

YI(O) = 1,Y2(0) = 0,Y3(0) = -1.


2. Find the approximating functions Y13, Y23, Y33, as in Problem 1, for thesystem

Y~ = Y2,Y~ = Yl,

Y~ = Y2 + Y3,

Yl(O) = 1,Y2(0) =' 0,Y3(0) = 2.

3. For what numbers b1, b2, b3 is there a solution (Yl, Y2, Y3) of the followingsystem such that Yl(l) = bl, Y2(1) = b2, Y3(1) = b3?

Y~ XVY2 + Y3/Yl,. Y~ = sin-1 Y3 + VYI -x,Y~ = x In (Yl -:- Y2).

4. Use Theorem 2 to prove Theorem 4.5. Write a first order system equivalent to

y" = xz' + y 2,

z" = Y' + xz.

What initial conditions can be prescribed for Y and z and their derivatives?6. Write a first order system equivalent to

y" z' + yy',z'" xz2 + Y' z" .

What initial conditions can be prescribed for Y and z and their derivatives?7. State and prove, using Theorem 1, an existence and uniqueness theorem

for the systemy" = F(x, Y, y', z, z'),z" = G(X,YiY',z,z').

8. Solve the system of Problem 2.9. Solve the following system by first eliminating z to obtain a third order

equation in y.y" 3y + z,z' -2y.

10. Solve the following system by first eliminating z.

y" z - y',z' 2z - 2y.

ANSWERS

1. Y13 = 1 ~ tx2 - ix3, Y23 = -x - tx2, Y33 = -1 - x + ix3

2. Y13 = 1 + tx2, Y23 = X - ix3, Y33 = 2 + 2x + ~X2 + !x3

3. 1 < bl, 0 < b2 < bl, Ib31 < 15. Y' = u, z' = v, u' = xv+ y2, v' = u+ xz; y(a), y'(a), z(a), z'(a) can be

prescribed arbitrarily for any a.


6. y' = u, z' = v, v' = w, u' = v+yu, vl = xz2 +uw; yea), y'(a), z(a),z'(a), z"(a) can be prescribed arbitrarily for any a.

7. If F, G, and their derivatives with respect to y, y', z, and z' are continuousin a region containing (a, b, c, d, e), then there is a unique solution (y, z) suchthat yea) = b, y' (a) = c, z(a) = d, z' (a) = e.

8. YI = cosh X, Y2 = sinh X, ya = eX + 19. y = cle-2x + c2ex + caxex, z = Cle-2x - 2c2ex + ca(-2x + 2)eX

10. y = clex+ c2e-V"2 X + cae-~ x,z = 2Clex + (2 + V2)c2e~" + (2 - V2)cae-~"

7-6 Polygonal approximations for systems. In this section we willgive a method for obtaining polygonal approximations to the two functionsY and Z which constitute the solution of the system

y' = F(x, Y, z),

Z' = G(x, Y, z),

yea) = b,

z(a) = c.(1)

Since the sequences {Yn} and {zn} of the existence proof converge to thesolution functions, we already have one method of obtaining approximatesolutions of (1). The polygonal approximations, however, are generallyeasier to calculate than the Picard approximations, ·even though the latterare more convenient to use in the proof itself.

With each partition {a = xo, XI, ... , Xn = d} of an interval [a, d)there is associated a polygonal approximation to each of the functions Yand z satisfying (1). Over the interval [a, Xl] the approximations are justthe lines tangent to the solution curves at (a, b) and (a, c); i.e., the linesthrough the points (a, b) and (a, c) with respective slopes F(a, b, c) andG(a, b, c). Let (XI, YI) and (XI, Zl) be the points at which the two tangentlines intersect the vertica,l line X = Xl' Compute F(XI' YI, Zl) andG(XI, YI, Zl), and take the lines through (Xl, YI) and (Xl, Zl) with theserespective slopes for the approximations over the interval [XI, X2]. Continuing in this way we obtain a sequence of points (a, b), (Xl, YI), ... ,(Xn, Yn), where the Yi are approximate values for Y(Xi) , and a similarsequence for the z-curve. The polygonal curves joining these vertices arethe approximations to Y and Z for the given subdivision. At each stagewe have (letting Yo = b, Zo = c)

Yk+l = Yk + F(Xk, Yk, Zk)(Xk+1 - Xk),(2)

Zk+l = Zk + G(Xk, Yk, Zk)(Xk+1 - Xk).

The method outlined above extends in the natural way to give polygonalapproximations to the functions in the solution of a first order systemin n functions. Since a second or higher order equation in one function

7-6] POLYGONAL APPROXIMATIONS FOR SYSTEMS 185

can be considered as a first order system, we can also use this method toobtain approximations to the solutions of such higher order equations.Consider, for example, the second order equation

y" = G(x, y, y'),

We rewrite this as the system

y(a) = b, y'(a) = c. (3)

y' = z,

z' = G(x, y, z),

y(a) = b,

z(a) = c,

and use (2), with F(x, y, z) = z, to compute simultaneous approximationsto y and y'.

EXAMPLE 1. y' = z - 2,z' = y + 3x,

y(O) = 1,z(O) = O.

We compute the approximations to y and z over the interval [0, 1), usingthe partition. {O, 0.2, 0.4, 0.6, 0.8, 1.0}. The calculations, to three decimal places, are shown in Table 1. The solutions of this system arey = eX - 3x, z = eX - 1. The approximate values from Table 1,rounded off to two decimal places, are listed' in Table 2 with the correctvalues for comparison. The graphs of the solutions and the approximations found are shown in Fig. 7-3.

EXAMPLE 2. y" = x + y, y(O) = 1, y'(O) = O. We treat this equation in the same way as the system y' = z, y(O) = 1; z' = x + y,z(Q) = O. Let us again compute approximations on [0, 1), using the partition {O, 0.2, 0.4, 0.6, 0.8, 1.0}. The calculations are listed in Table 3.The correct solutions, y = eX - x, z = y' = eX - 1, and the approximations found above are shown in Fig. 7--4.

TABLE 1

Xk 0 0.2 0.4 0.6 I 0.8 I 1.0Yk 1 0.6 0.24 -0.072 -0.326 -0.511Zk 0 0.2 0.44 0.728 1.074 1.489

Zk - 2 -2 -1.8 -1.56 -1.272 -0.926!!.Yk = !(Zk - 2) -0.4 ·-0.36 -0.312 -0.254 -0.185Yk + !!.Yk = Yk+l 0.6 0.24 -0.072 -0.326 -0.511

Yk + 3x 1 1.2 1.44 1.728 2.074!!.Zk = t(Yk + 3x) 0.2 0.24 0.288 0.346 0.415

Zk + !!.Zk = Zk+l 0.2 0.44 0.728 1.074 1.489


TABLE 2

[CHAP. 7

x 0 0.2 0.4 0.6 0.8 1.0Yk 1 0.60 0.24 -0.07 -0.33 -0.51

Y = e" - 3x 1 0.62 0.29 0.02 -0.17 -0.28Zk 0 0.20 0.44 0.73 1.07 1.49

z=e"-1 0 0.22 0.49 0.82 1.23 1.72

TABLE 3

Xk 0 0.2 0.4 0.6 0.8 1.0Yk 1 1.0 1.04 1.128 1.274 1.489Zk 0 0.2 0.44 0.728 1.074 1.489

AYk = 1,-Zk 0 0.04 0.088 0.146 0.215Yk + AYk = Yk+l 1 1.04 1.128 1.274 1.489

Xk + Yk 1 1.2 1.44 1.728 2.074AZk = 1,-(Xk + Yk) 0.2 0.24 0.288 0.346 0.415

Zk + AZk = Zk+l 0.2 0.44 0.728 1.074 1.489

-0.5

FIG. 7-3. Polygonal approximationsfor Y' = Z - 2, z' = .y + 3z, yeO) = 1,z(O) = O.

FIG. 7-4. Polygonal approximationsfor y" = X + Y, yeO) = 1, y'(O) = O.

7-6] POLYGONAL APPROXIMATIONS FOR SYSTEMS

PROBLEMS

187

1. Compute approximations to the solutions of Example 1 using the partition{O, 0.1, 0.2, ... , 0.9, 1.0 j. Compare your results with the correct values inTable 2.

2. Using the following partitions, find an approximation to the solution of theequation of Example 2.

(a) 10, 0.1, 0.2, ... , 0.9, 1.0 j of [0, 1],(b) {-0.5, -0.4, -0.3, -0.2, -0.1, OJ of [-1,0].

3. For the system Y' = !Z, z' = -ty, yeO) = !, z(O) 1, find approxima-tions to y and z for the partition {O, 0.2, 0.4, 0.6, 0.8, 1.0 j. Show that thesolution is y = !eX

, z = e-x , and plot the approximations and solution curveson the same graph.

4. Find approximations to the solution curves of

y' = 2(1 - z)/x2 ,

Z' = -3y, z(l)

y(l)

0,

1,

(a) for the partition {0.2, 0.4, 0.6, 0.8, 1.0},(b) for the partition {0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0j.

Verify that the solution is y = x2, Z = 1 - x3, and plot these curves with theapproximations.

5. Plot approximations to the solution curves of the system of Problem 4 byusing the partition 10.1, 0.2, 0.3, 0.4, 0.51 of [lo, !l-

(a) Start with the initial values ym = 0.250, zm = 0.875.(b) Start with the initial values y(lo) = 0.010, z(lo) = 0.999.

6. Graph the polygonal approximations to y and y', where y is the solutionof y" + y = 0, yeO) = 0, y'(O) = 1, for the partition

{O, 0.1, 0.2, ... , 1.5, 1.6, 1.71

of [0, 1.7]. What are the approximate values obtained for Y('/T/2)? for y'(7r/2)?

ANSWERS

1. Yk: 1, 0.800, 0.610, 0.431, 0.264, 0.110, -0.029, -0.152, -0.257, -0.343,-0.407,

Zk: 0, 0.100, 0.210, 0.331, 0.464, 0.610, 0.771, 0.948, 1.143, 1.357, 1.5932. (a) Yk: 1, 1, 1.010, 1.031, 1.064, 1.110, 1.171, 1.248, 1.343, 1.457, 1.593,

Zk: 0, 0.100, 0.210, 0.331, 0.464, 0.610, 0.771, 0.948, 1.143, 1.357, 1.593(b) Yk: 1, 1, 1.010, 1.029, 1.056, 1.090,

Zk: 0, -0.100, -0.190, -0.271, -0.344, -0.4103. Yk: 0.500,0.600,0.725,0.883, 1.085, 1.347,

Zk: 1.000, 0.800, 0.633, 0.495, 0.382, 0.290


4. (a) Yk: 1, 0.600, 0.350, 0.306, 0.731Zk: 0, 0.600, 0.960, 1.170, 1.354

(b) Yk: 1, 0.800, 0.627, 0.483, 0.372, 0.301, 0.289, 0.383, 0.743,Zk: 0,0.300,0.540,0.728,0.873,0.985, 1.075, 1.162, 1.277

5. (a) Yk: 0.250, 0.150, 0.087, 0.076, 0.181,Zk: 0.875, 0.950, 0.995, 1.021, 1.044

(b) Yk: 0.010, 0.030, 0.050, 0.079, 0.114,Zk: 0.999,0.996,0.987,0.972,0.948

6. Yk: 0, 0.100, 0.200, 0.299, 0.396, 0.490, 0.580, 0.665, 0.744, 0.817, 0.882,0.939, 0.987, 1.026, 1.055, 1.074, 1.082, 1.079,

Zk: 1, 1, 0.990, 0.970, 0.940, 0.900, 0.851, 0.793, 0.726, 0.652, 0.570, 0.482,0.388,0.289,0.186,0.080, -0.027, -0.135At 7r/2, v-approximation is 1.080, v'-approximation is 0.004.

7-7 Linear systems. We return now to the linear system (4), Section 7-5, and indicate how the theory for such a system parallels that fora single linear equation (cf. Section 3-3). The system in question is

y'l = PllYI + ... + PlnYn + qI,(1)

y~ = PnlYI + ... + PnnYn + qn,

where the functions Pij and qi are assumed continuous on some interval[a, (j). We know (Theorem 2, Section 7-5) that there is exactly one solution (YI, ... , Yn) of (1) on [a, (j] such that YI(a) = bl , ... , Yn(a) = bnfor any a in (a, (j) and any numbersbI, ... ,bn . If the functions qi areall zero, then the system (1) is called homogeneous. The homogeneoussystem

yi·= PIIYI +.,. + PlnYn,

(2)

y~ = PnlYI + ... + PnnYn

is called the reduced form of (1).We define the sum of two solutions of (1) or (2), and a constant multiple

of a solution as follows:

(Yll, ... , YIn) + (Y21, ... , Y2n) = (Yll + Y2I, ... ,YIn + Y2n),

I t then makes sense to talk of linear combinations

of solutions. A set of solutions is linearly independent if no nontrivial

7-7] LINEAR SYSTEMS 189

linear combination is equal to the n-tuple (0, ... ,0). The Wronskian ofn solutions of (1) or (2) is the n-by-n determinant whose columns are thegiven solutions:

W(X)

Yll(X) Y21(X)

YI2(X) Y22(X)

Yln(X) Y2n(X) Ynn(X)

With the above definitions for sum, linear combination, Wronskian, etc.,all the theorems of Section 3-3 hold for the system (1). To solve (1), therefore, it is necessary and sufficient to find one particular solution of (1),and all solutions of the reduced system (2). A linear combination of solu~ions of (2) is again a solution of (2). Any n solutions of (2) are linearlyindependent if and only if their Wronskian never vanishes. The set of alllinear combinations of any n linearly independent solutions of (2) is thefamily of all solutions of (2). If (YOI' ,Yon) is a particular solutionof (1), and (Yll, . .. , YIn), ... , (Ynl, , Ynn) are linearly independentsolutions of (2), then every solution (Yb , Yn) of (1) can be expressedas follows.

YI = YOI + CIYll + + CnYnl,

Y2 = Y02 + CIYl2 + + CnYn2,

Yn = YOn + CIYln + ... + CnYnn.

(3)

Now we turn to the case in which the coefficients Pij are constants,since this is the only case in which we have a systematic way of findingsolutions to specific problems. We consider the homogeneous system (2),and illustrate the methods of finding solutions for the case n = 2:

y' = PIIY + P12z,(4)

z' = P2lY + P22Z.

We must find two linearly independent solutions (YI, Zl), (Y2' Z2), and byanalogy with a single homogeneous linear equation, we try to find solutionsof the form (AeTX, BeTX ). Substitution of Y = AeTX , Z = BeTX in (4) givesthe equations

AreTX = PllAeTx + PI2BeTx,

BreTX = P2lAeTx + P22BeTX.

Hence (AeTX, BeTX ) is a solution of (4) if the numbers A, B, and r satisfy


the following simultaneous equations:

(pu - r)A + P12B = 0,

P2lA + (P22 - r)B = O.(5)

(6)

This system of linear equations in A and B has a nontrivial solution ifand only if the determinant equals zero; i.e., if the number r satisfies theequation:

IPu- r Pl2 1=0.

P21 P22 - r

The quadratic equation (6) is called the auxihary equation for the system(4). Suppose rl and r2 are distinct roots of (6), and AI, B I are numberswhich satisfy (5) when r = rb and A 2 , B 2 are numbers which satisfy (5)when r = r2. Then (Aler1X, Bler1X ) and (A 2er2X, B 2er2X ) are solutions of(4), and every solution (y, z) of (4) can be written

y = clAlerix + c2A2er2x,

z = clBlerlx + c2B2er2x.

If the auxiliary equation has only one root ro, then two solutions can befound by trying functions of the form y = (AI + A 2x)erox , z =(B I + B 2x)erox. If the roots of (6) are the complex numbers ro and 1'0,then there will be complex numbers A o and B o satisfying (5) when r = ro.The pair (Aoerox, Boerox ) is a complex solution, and the real and imaginaryparts are two real solutions.

The methods outlined above also extend to homogeneous systems inthree or more functions (see Example 4).

EXAMPLE 1. y' = y + 2z,

The auxiliary equation is

z' = 2y + Z.

1

1 - r 2 I = r2 - 2r - 3 = 0,2 1 - r

with roots r = 3, and r = -1. For r = -1 the equations (5) are both2A + 2B = 0, so we must have A = -B. We can take, for example,A = 1 and B = -1, and (e-X

, -e-X) is one solution of the system.

For r = 3 the equations (5) are both equivalent to A = B. Hence wehave the second solution (e 3x, e3x). Any solution (y, z) can therefore beexpressed

y = cle-x + C2e3x,

z = -cle-x + C2e3x.

7-71 LINEAR SYSTEMS 191

EXAMPLE 2. y' = 3y - 2z, z' = 2y - z.

The auxiliary equation for this system is

-2 I 2= (r - 1) = 0,-1 - r

with the single root r = 1. Here we try solutions y = (Ai + A 2x)eX,

z = (B l + B 2x)ex. Substituting these functions in the system and divid

ing out eX gives·

Ai + A 2 + A 2x = 3A l - 2Bl + (3A 2 - 2B2)x,

B l + B 2 + B 2x = 2A l - B l + (2A 2 - B 2)x.

For these equations to be identities, we must have

Ai + A 2 = 3A l - 2B l ,

A 2 = 3A 2 - 2B2,

B l + B 2 = 2A l - B l ,

B 2 = 2A2 - B 2.

The second and fourth equations reduce to A 2 = B 2 , and the first andthird equations both become 2A l - A 2 - 2B l = 0. We can take, forexample, Ai = B l = 1, A 2 = B 2 = 0, or Ai = 1, B l = 0, A 2 =B 2 = 2. The solutions obtained for these values are (eX, eX) and[(1 + 2x)eX, 2xeX]. All solutions of the system can be expressed

Y = clex + c2(1 + 2x)eX = (Cl + C2 + 2c2x)eX,

z = clex + C22xex = (Cl + 2c2x)ex.

EXAMPLE 3. Y' = Y - z, z' = y + z.

The auxiliary equation is r 2 - 2r + 2 = 0, with the complex roots1 + i, 1 - 1. For r = 1 + i the equations (5) both read

[1 - (1 + i)]A - B = 0,

or A = iB. Taking B = 1, A = i, we get the complex solution(ieO+ ilx, e(l+ilx), or

(ieXcos x - eX sin x, eX cos x + iex sin x).

The real parts and imaginary parts must also be solutions, and we have

192

the two real solutions

SYSTEMS OF EQUATIONS [CHAP. 7

(-eX sin x, eX cos x),

The family of all solutions is

(eX cos x, eX sin x).

y = -Clex sin x + C2ex cos X,

Z = Clex cos x + C2ex sin x.

If the other root, r = 1 - i, of the auxiliary equation is used, one arrivesat the same pair of solutions.

EXAMPLE 4. To illustrate how these methods extend to three equationsin three functions, we consider the system

y' = y+w,

Z' = 2y - z,

w' = 2w.

There are solutions of the form (Aerx, Berx, Cerx ) for numbers r whichsatisfy

1 - r

2

o

o-1 - r

o

1

o2 - r

-(1 - r)(1 + r)(2 - r) = O.

We treat the roots 1, -1,2 in turn. The triple (AeX, Bex, CeX) is a solution of the system if

OA + OB + 1C = 0,

2A - 2B + OC = 0,

OA + OB + 1C = O.

These equations are satisfied by A = B = 1, C = 0, and (eX, eX, 0) is asolution. Similarly, if we substitute the functions y = Ae-x

, z = Be-X,w = Ce-x in the system, we get the equations

2A + OB + lC = 0,

2A + OB + OC = 0,

OA + OB + 3C = 0,

which are satisfied by A = C = 0, B = 1. Thus (0, e-x , 0) is a second

7-7] LINEAR SYSTEMS 193

solution of the system. For r = 2, the condition that (Ae 2z, Be2z, Ce2z)be a solution is

(-1)A + OB + 1e = 0,

2A - 3B + oe = 0,

OA + OB + oe = o.

We must have e = A, and B = ~A; taking A = 3 we get the thirdsolution (3e2z, 2e2z, 3e2z). Any solution (y, z, w) can be expressed as alinear combination of (eZ, eZ, 0), (0, e-z, 0), and (3e2z, 2e2z, 3e2Z), and wecan express the solutions

PROBLEMS


1. y' 2y + 3z, 2. y' = y, 3. y' 2y + 5z,z' 2y+ z z' = -2z z' y - 2z

4. y' y, 5. y' =y-z 6. y' y - z,z' 2y+ z z' = y+ 3z z' 2y - z

7. y' y - 2z, 8. y' = y+ 2w,z' = y+ 3z z' = y+ 2z+ w,

w' = 3y9. y' = 2y - 2z - 4w,

z' = 2y - 3z - 2w,w' = 4y - 2z - 6w. The auxiliary equation is -(r + 2)2(r + 3) = O.

Show that there are three linearly independent solutions of the form

(Ale-2%, Ble-2%, Cle-2%), (A2e-2%, B2e-2%, C2e-2%),(A3e-3%, B3e-3%, C3e-3%).

ANSWERS

1. cle-% + 3c2e4%, z = -Cle-% + 2c2e4%2. y Cle%, z = c2e-2%3. y = 5Cle3%- C2e-3%, z = Cle3%+ c2e-3%4. y = cle%, z = cle% + 2c2xe%5. y cle2

% + C2xe2%, z = -clez%- c2(l + x)e2%6. y Cl cos x + C2 sin x, z = Cl (cos x + sin x) + C2 (sin x - cos x)7. y -2cle2%cos x - 2c2e2%sin x

z cle2%(cos x - sin x) + c2e2%(cos x + sin x)8. y = c2e3%+ 8c3e-2%, z = cle2%+ 2c2e3%+ C3e-2%, w = c2e3%- 12c3e-2%9. (0, -2e-2%, e-2%), (e-2%, 2e-2%, 0), (2e-3%, e-3%, 2e-3%);

y = c2e-2%+ 2c3e-3%,z = -2Cle-2%+ 2c2e-2%+ C3e-3%,

w = Cle-2%+ 2c3e-3%


7-8 Operator methods. We have so.far considered only systems of equations of the simple form in which each derivative is expressed explicitlyin terms of the unknowns. It is frequently necessary to consider moregeneral systems, in which the derivatives of several functions occur ineach equation, and second or higher order derivatives appear. An exampleof such a system is

2y" - 4y - z'·- 4x,

2y' + 4zi- 3z. 0,

which can be conveniently written in operator notation as

(2D 2- 4)y - Dz = 4x,

(1)2Dy + (4D - 3)z = O.

We consider systems, such as (1) above, which are linear in the unknownsand their derivatives and have constant coefficients. The general form ofsuch a system is

L ll (D)Yl + + L1n(D)Yn = qI,

L 21 (D)Yl + + L 2n (D)Yn = q2,

(2)

where the Lij(D) are linear operators with constant coefficients, and theqi are arbitrary continuous functions.

Since our existence theorem (Theorem 2, Section 7-5) is stated forlinear systems in the simple form (4), Section 7-5, which we will call thebasic form, we ask when the system (2) is equivalent to a system of thisbasic form. If we can solve (2) algebraically for the highest order derivatives of Yl, ... , Yn which occur in the system, we can write an equivalentsystem in the basic form by introducing the lower order derivatives asnew unknowns. To illustrate with the system (1), consider this as a linearalgebrat"c system in D 2y and Dz:

2D2y - Dz = 4y + 4x,(3)

4Dz = -2Dy + 3z.

The determinant of this system is

102

-411 = 8 -,t. 0,

7-8] OPERATOR METHODS 195

so we can solve for D 2y and Dz, and we get

D 2y = -!Dy + 2y + iz + 2x,

Dz = -~Dy + iz.

If we let Dy = w, whence D 2y = Dw, we obtain the following systemin basic form which is equivalent to (1):

Dy = w,

Dw = -!w + 2y + iz + 2x,

Dz = -~+ iz.

(4)

It follows that (1) has solutions and that arbitrary initial conditions canbe prescribed for y, Dy, and z. The general solution of (4) will containlinear combinations of three linearly independent solutions of the reducedsystem. Therefore the general solution of (1) or (3) will have y and zexpressed as linear combinations involving three arbitrary constants.

If the system (2) cannot be solved algebraically for the highest orderderivatives which occur, it is not equivalent to a system in basic form andmay in fact be inconsistent. For example, the following system is inconsistent.

(D 2 + 2D)y + (2D - 3)z = eX,

(D 2 + 2D)y + (2D - 3)z = O.

We will treat only those systems which can be put in the basic form, theso-called nondegenerate systems.

Our methods of solving systems such as (1) frequently introduce extraneous functions. That is, we obtain formulas for y and z which arenecessary conditions that (y, z) be a solution, but not sufficient. In practice, this means that the formulas contain more constants than theyshould, and it is convenient to have a check on how many arbitraryconstants should appear. There will be as many arbitrary constants asthere are functions in the equivalent basic system, which introduces a newfunction for each derivative occurring in the original system except thoseof highest order. For example, suppose that D 3y and D 2z are the highestorder derivatives which appear in a system of two equations in y and z.

_ The equivalent basic system involves the five functions y, Y2 = Dy,Y3 = D 2y, Z, Z2 = Dz, and there must be five arbitrary constants in thesolution. In general, the number of constants in the solution is the sum of thehighest orders of the derivatives which occur in the system.

The preceding discussion indicates generally what sort of solutions thesystem (2) can be expected to have. Now we turn to the problem of

196 SYSTEMS OF .EQUATIONS [CHAP. 7

finding the solutions. We proceed in much the same way as with algebraicsystems. From the given system, we obtain new equations by differentiating and multiplying by constants, etc., so these -equations can be combined to yield an equation in just one of the unknowns. This equation issolved and the solution substituted in the system. The procedure isillustrated in the examples below.

EXAMPLE 1.y' = -y + z,

(5)z' = -5y + 3z.

This system can be written in operator form

(D + l)y - z = 0,

5y + (D - 3)z = O.

Operating on (6) with (D - 3), we obtain

(D - 3)(D + l)y - (D - 3)z = O.

Adding (8) and (7), and simplifying, we get

[(D - 3)(D + 1) + 5]y = 0,

(D 2- 2D + 2)y = O.


From (6), z = (D + l)y,or

z = Cle'" (2 cos x - sinx) + c2e"'(2 sin x + C08X).

(6)

(7)

(8)

(9)

(10)

Here the Eqs. (10) and (6) form a system equivalent (Problem 1) to theoriginal system (6) and (7), and there is no problem of extraneous solutions. Note that the formulas for y and z contain two constants, as theyshould.

EXAMPLE 2.(D - l)y + (D + l)z = e-"',

(11)

To eliminate z, operate on the first equation with D and on the second

7-8]

with (D + 1) to obtain

OPERATOR METHODS 197

D(D - l)y + D(D + l)z = -e-x ,

(D + 1) D 2y + (D + 1) Dz = O.

Subtracting the corresponding sides of equations (12), we get

[(D + I)D 2 - D(D - 1)]y = e-x ,

(Da + D)y = D(D2 + l)y = e-x•


From the second of equations (11), we have

Dz = 2e-x - D 2y = !e-X + C2 cos X + Ca sin x,

z = -!e-X + C2 sin x - Ca cos x + C4.

(12)

(13)

(14)

(15)

Since D 2y and Dz are the highest order derivatives which appear, thereshould be 2 + 1 = 3 arbitrary constants instead of four. The functionsy and z of (14) and (15) satisfy the second equation of (11) for all valuesof Cl, C2, Ca, and C4. However, substitution in the first equation of (11)shows that (y, z) is a solution if and only if Ci = C4. Hence the solutionsof (11) are given by

y = Cl + C2 cos X + Ca sin x - -!e-x,

z = Cl + C2 sin x - Ca cos x - !e-x•

EXAMPLE 3. We solve the system (1). Operating on the first equationwith (4D - 3), and on the second with D, we get

(4D - 3)(2D2 - 4)y - (4D - 3) Dz = 16 - 12x,

2 D 2y + D(4D - 3)z = O.

_ Adding these equations and simplifying, we obtain

2[(4D - 3)(D2 - 2) + D 2]y = 16 - 12x,

(2Da - D 2 - 4D + 3)y = 4 - 3x,

(2D + 3)(D - 1)2y = 4 - 3x.


The solutions of this equation are

y = -x + Cle'" + C2xe'" + C3e-<3/2)",.

From the first equation of (1), we have

Dz = (2D 2 - 4)y - 4x

-2Cle'" + C2(-2x + 4)e'" + !C3e-<3/2)",.

Hence z must satisfy

z = -2cle'" + C2( -2x + 6)e'" - -!C3e-<3/2)", + C4.

Substituting y and z in the second of equations (1) gives 3C4 + 2 = 0,so C4 = -~. Accordingly the solutions are given by

y = -x + cle'" + C2xe'" + C3e-<3/2)""

z = -~ - 2cle'" + C2(-2x + 6)e'" - -!C3e-<3/2)",.

PROBLEMS

1. Show that the system consisting of equations (10) and (6) is equivalent tothe system of equations (6) and (7). [Since (10) was derived from (6) and (7),all that remains is to assume (10) and (6), and derive (7)].

2. Solve the system (5) of Example 1 by means of the auxiliary equation, asin Section 7-7.

3. Substitute the formulas (14) and (15) in the system (11) of Example 2and show that C1 = C4 is necessary for (y, z) to be a solution.

4. Show that the following system is nondegenerate. How many constantsmust appear in the general solution?

D2y + (D + l)z + Dw = eX,

(2D + l)y + 2z + D2W = 1,

(D3 + D2)y + (D + 2)z + (D - 3)w = x.

5. (See Example 3.) Show that (y, z, w) = (-x, -~, -1) is a particularsolution of the basic system (4) which is equivalent to (1). Show that

7. D2y + (2D + 3)z 3x,Dy+Dz=x-l

and

are solutions of the reduced form of (4).


6. (D - l)y + z = ex,-2y + (D + l)z = 1

7-9] LAPLACE TRANSFORM METHODS 199

8. (D - I)y - Dz = eX,2Dy - (D + I)z = 2eX

9. (D - I)y + (D2 - I)zD2y + (D + I)z = 0

10. Show that the following system is degenerate and that there is exactlyone pair of functions which satisfies the system

(D2 - 2)y + Dz 2x,

3Dy + 3z = 1.

11. Show that the following system is inconsistent.

(D2 - D)y + (D + I)z + (D3 - D2)W eX,

(D + I)y + (D - I)z - (D3 + D)w = 1,

(D2 + I)y + 2Dz - (D2 + D)w = 2x.

Show that the system is consistent if the right member of the last equation ischanged to eX + 1. What can you say about the solutions?

ANSWERS

2. (e(l+i)x, (2 + i)e(l+i)x) is a complex solution.4. six6. y = !Cl (cos X - sin x) + !C2 (cos X+ sin x) + eX - 1,

z = Cl cos x + C2 sin x + eX - 1 ..7. Y tx2 - 2x - Cle3x - c2e-x + C3,

Z = x-I + Cle3x + c2e-x

8. y = Cl cos x + C2 sin x,z = Cl (cos X - sin x) + C2 (sin x + cos x) - eX

9. y = _e2x + Cle-x + c2ex + ~3xex,z = !e2x - clxe-x - !C2eX - C3(!X + £)eX+ c4e-x

10. y = -x, z = !11. If w is any function, there is a three-parameter family of solutions for y

and z in terms of w.

7-9 Laplace transform methods. The Laplace transform provides aneffective way of treating systems of linear equations with constant coefficients, as well as single equations in one unknown. By taking transforms,we convert a system of linear equations in y and z into an algebraic systemof linear equations in y and z. If the original system is consistent, .thisalgebraic system can be solved for y and z, and then yand z can be determined by the methods of Chapter 5. As for a single equation, the Laplacetransform method gives directly the particular solution which correspondsto a given ,set of initial conditions.

In this section we will illustrate how the transform method applies tosome of the examples of the precel1ing sections.


EXAMPLE 1. (cL Example 1, Section 7-7.) We consider the system

y' = y + 2z,(1)

z' = 2y + z, '

with the initial conditions

yeO) = 2, z(O) = O. (2)

Taking transforms of both sides of equations (1), we get

8'0 - 2 = '0 + 22,

82 - 0 = 2'0 + 2,

or(8 - 1)'0 - 22 = 2,

-2'0 + (8 - 1)2 = o.

The determinant of the system (3) is

1

8- 1 -2 1 2= 8 - 28 - 3 = (8 - 3)(8 + 1).-2 8 - 1

Hence '0 and 2 are given by

(3)

21 4o = (8 - 3)(8 + 1) .

~1~12 21'0 = (8 - 3)(8 + 1) o 8 - 1

2 = 7(8-~37"C{(;-8 ~+-1=) 18_/

28 - 2= (8 - 3)(8 + 1) ,

From the last equation above and #7 of Table 2 (Section 5-4), we get

Z= 4 [ 3% _ -%] _ 3% _ -%

3.- (-1) e e - e . e .

Using #8 and #7 of Table 2 in Section 5-4, and the formula above for '0,we get

y = f[3e3% + e-%] - f[e3

% - e-%]

= e3%+ e-%.

If, instead of the initial conditions (2), we take arbitrary initial conditions yeO) = a, z(O) = b, then the method above gives the general


solution of the system (1) in the form (Problem 1)

z = -!(a + b)e3z - !(a - b)e~z.

EXAMPLE 2. (cf. Example 4, Section 7-7.)

(4)

y' = y + w,

z' = 2y - z,

w' = 2w,

yeO) = 4,

z(O) = 2,

w(O) = 3.

(5)

The transformed equations for the system (5) are

8'0 - 4 = '0 + '111,

82 - 2 = 2'0 - 2,

sW - 3 = 2'111,

which can be written

(8 - 1)'0 + 02 - 1'111 = 4,

-2'0 + (8 + 1)2 + OW = 2,

0'0 + 02 + (8 - 2)'111 = 3.

From the last of equations (6), we get

A 3w=--·

8-2

Substitution of this result in the first equation gives

A 4 3 1 3y = --+ = --+-_.

8 - 1 (8 - 2)(8 - 1) 8 - 1 8 - 2

From the second of equations (6), we find that

2 = _2_+ 2 + 68 + 1 (8 - 1)(8 + 1) (8 - 2)(8 + 1)

Decomposing the terms above into partial fractions gives

2=_1 1_+_2_.8-1 8+1 8-2

(6)

202 SYSTEMS: OF EQUATIONS [CHAP. 7

With the formulas above forw, '0, and i, we can identify the solution as

y = eX + 3e2x,

z = eX - e-X+ 2e2x,

w = 3e2x•

The next example shows how the transform method applies to the higherorder systems treated in Section 7-8. Notice how the operator manipulations of Example 2, Section 7-8, parallel the algebraic operations in thefollowing example.

EXAMPLE 3. (cf. Example 2, Section 7-8.)

y' - y + z' + z = e-x,

y" + z' = 2e-x,

yeO) = 2,

z(O) = -1

y'(O) = t(7)

The equations in '0 and i corresponding to (7) are

BY - 2 - '0 + 8i + 1 + i = _1_,8+1

82•1 - 28 - 1. + 8i + 1 = _2_,Y 2 8 + 1

which we write as

(8 - 1)'0 + (8 + l)i = 8 ~ 1 + 1 = : ~ ~ '

(8)

To solve for '0 and i, multiply the first of equations (8) by 8, and thesecond by 8 + 1; this gives

(. • _ ._ 82 + 28

8 S - l)y + 8(8 + l)i = 8 + 1 '

2, • 1 483 + 782 + 68 + 38.(8 +_l)y +.8(8-+ l)z.- 2 - - -8-+ ·1-- - .

Subtracting the corresponding sides of (9), we get

( 3 + )' _ (2 + 1) '_ .! 483

+ 582

+ 28 + 38 8Y-88 y-2 8+1 '

(9)


or

Hence

1 483 + 58

2 + 28 +3! t 8 (10)Y = 2 8(8 + 1)(82 + 1) = S - 8 + 1 + 82 + 1 .

Y = i - te-x + cos x,

Z=

which is Formula (14), Section 7-8, with Cl = -i, C2 = 1, and C3 = O.From the second of equations (8), and (10), we obtain

.+2+ 2 t-sy 8(8 + 1) - S

1 483 + 382 + 28 + 3 2- 2 (8 + 1) (82 + 1) + 2 + 8(8 + 1)

1 82- 28 - 1 2 _ ]:.

- 2 (8 + 1)(82 + 1) + 8(8 + 1) 8

Breaking the terms above iflto partial fractions, we get

. -t 1 2 2 tz = 8 + 1 + 82 + 1 + s - 8 + 1 - s

_ -t +i+_1_.- 8 + 1 8 82 + 1

Hencez = --ie-x + i + sin x,

which is Eq. (15), Section 7-8, again with C2 = 1, and C3 = O.

PROBLEMS

1. Use transforms, with the initial conditions yeO) = a, z(O) = b, to obtainthe general solution (4) of the system (1).

2. Use transforms to solve the systems cited, for the given initial conditions:

. (a) System of Example 2, Section 7-7; yeO) = 0, z(O) = -1(b) System of Example 3, Section 7-7; yeO) = 1, z(O) = 0(c) System of Example 1, Section 7-8; yeO) = 1, z(O) = 2Cd) Problem 6, Section 7-8; yeO) = 0, z(O) = 0(e) Problem 7, Section 7-8; yeO) = 1, z(O) = -1(f) Problem 8, Section 7-8; y(O) = 0, z(O) = 0

_ (g) Problem ll, Section 7-8; y.(O) .= =1, z(O)-= !3. THEOREM: For any transform!, lim.->oo f(s) = o.

Use this fact to show that the system of Problem 10, Section 7-8, has only onesolution. [Hint: Let yeO) = a, y'(O) = b, z(O) = c be any initial conditions;show that y = -1/s2 + ill - 3b - 3c], and hence that y = -x, etc.]


ANSWERS

2. (a) y = 2xe"', x = -1 + 2xe'"(b) y e'" cos x, z = e'" sin x(c) y e'" cos x, z = e"'(2 cos x - sin x)(d) y e'" - 1, Z = e'" - 1(e) y -!x2 - 2x + 1, Z = x-I(f) y sin x, z = sin x + cos x - e'"(g) y -e2"', z = !e2'"

INDEX

Absolute convergence, 49Analytic function, 50Analytic solution, 53Approximations,

Picard, 144for systems, 172, 180

polygonal, 156-160for systems, 184-187

Arbitrary constants, 195Auxiliary equation, 68, 86

for a linear system, 190

Basic form of a linear system, 194Bernoulli equation, 45

Catenary, 166Cauchy's existence proof, 156Clairaut equation, 20Coefficients, undetermined, 100Comparison test,

for improper integrals, 116for power series, 51for series of functions, 142

Complex conjugate, 87 (Prob. 6)Complex exponential, 84Complex function, 83Complex solution, 85Connected set, 37Continuous function, 140Convergence,

of improper integrals, 116of power series, 48of a sequence of functions, 141, 142of a series of functions, 142

Convolution, 136Critical damping, 76Curve, 35Curves, families of, see Families of

curves

D,88Damping, 75, 76

Derivative of a complex function, 83Determinant- of a system of equations,

60Differential equations,

Bernoulli, 45Clairaut, 20equivalent, 2exact, 6, 31of families of curves, 15homogeneous, 25linear, see Linear differential

equationsorder of, 1ordinary, 1partial, 1solution of, 2solvable for y, 23systems of, see Systems of differen-

tial equationsDifferential operators, 88-94Distance between complex numbers, 83Domain, 37

simply connected, 38

Electric circuits, 74Envelopes, 17Equations solvable for y, 23Equivalent equations, 1Exact equations, 6, 31Existence and uniqueness theorem,

for first order equations, 12, 155for first order systems, 177for linear equations, 58, 179 (Prob.

10), 182for linear systems, 178, 180for nth order equations; 181

Exponential order, 124

Families of curves, 15differential equations of, 16envelopes of, 17

205

206 INDEX

orthogonal, 46self-orthogonal, 47, 48 (Prob. 8)

Flexible cable, 166Frequency, 76

natural damped, 79Fundamental theorem of calculus, 140

General solution, 66Green's theorem, 39 (Prob. 6)

Homogeneous algebraic system, 60Homogeneous first order differential

equation, 26Homogeneous function, 25Homogeneous linear differential equa

tion,95Homogeneous system of linear differ

ential equations, 188

Imaginary part, 83Implicitly defined solutions, 4Improper integrals, 116

uniformly convergent, 117Integral form of a differential equa

tion, 144Integral inequalities, 141Integrating factor,

depending on x only, 143for first order linear equations, 4l

Interval of convergence, 48Inverse operators, 105, 130

£,119£-1,130Laplace transforms,

defined, 119properties of, 124-131in systems, 199table of, 132

Line integrals, 35Linear algebraic equations, 59

homogeneous, 60Linear dependence and independence,

64Linear differential equations,

existence theorem for, 58, 179 (Prob.10), 182

first order, 40general solution of, 66homogeneous, 58particular solution of, 66reduced form of, 61second order, with constant coeffi

cients, 68-73Linear differential operator, 88Linear systems,

auxiliary equation for, 190basic form of, 194homogeneous, 188Laplace transform in, 199nondegenerate, 195reduced form, 188theory of, 188in two functions, 177

Lipschitz condition, 149, 173

Method of undetermined coefficients,70,100

Method of variation of parameters, 109Motion, of a particle, 165

simple harmonic, 76, 81M-test, 117

Natural damped frequency, 79Numerical methods, see Approxima

tions

Object function, 119Open set, 37Operator methods, 88-94

for systems, 194-199Operators, 88

inverse, 105sum and product of, 89

Order of a differential equation, 1Orientation of a curve, 36Orthogonal families, 46Overdamped system, 75

Parametric equations, 35, 168of the catenary, 167

Parameters, variation of, 109Partial fractions, 107, 134, 135Particle, motion of, 165

INDEX 207

Particular solution, 66Pendulum, 79Period of a vibration, 76Picard's method,

outline of, 144-147for systems, 171-175

Picard's theorem, 155Pointwise convergence, 141Polygonal approximations, 157

for systems, 184Power series, review of, 48-52

solutions, 53-57

Radius of convergence, 48Ratio test, 51Real part, 83Reduced equation, 61, 188Reduction of order, 114 (Prob. 9)Resonance, 79

Separation of variables, 3-7Sequences and series of functions, 141,

142Series, power, 48-52

solutions in, 53-57Simple harmonic motion, 76, 81Simply connected domain, 38Simultaneous equations,- see SystemsSolution, 1

general,66particular, 66of a system, 161

Spring-mass systems, 74Substitution, 26Successive approximations, see Picard's

method

Superposition principle, 62System of differential equations, 161

199equivalent to a first order equation,

168linear, see Linear systemspolygonal approximations for, 184solution of, 161

Table of Laplace transforms, 132Tangent field, 9

for a system, 162Trajectory, orthogonal, 46Transform, 119Transients, 75

Underdamped system, 76Undetermined coefficients, method of,

70, 100Uniform convergence,

of improper integrals, 117of power series, 49of sequences of functions, 142

Uniqueness theorems, see Exist~~e

- and uniqueness theorems

Variables separate, 3-7Variation of parameters, 109

Weierstrass M-test, for integrals, 117Weierstrass polynomial approximation

theorem, 128Work, 36Wronskian, 64, 189

Differential Equations

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Transcript of Differential Equations